Question 13 Marks
The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive $ ^{14}_6\text{C}$ present with the stable carbon isotope $^{12}_6\text{C}.$ When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of $^{14}_6\text{C },$ and the measured activity, the age of the specimen can be approximately estimated. This is the principle of $^{14}_6\text{C}$ datingused in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
AnswerDecay rate of living carbon-containing matter, R = 15 decay/min
Let N be the number of radioactive atoms present in a normal carbon- containing matter.
Half life of $^{14}_6\text{C },\ \text{T}_{1/2}=5730\text{ Years}$
The decay rate of the specimen obtained from the Mohenjodaro site:
R' = 9 decays/min
Let N' be the number of radioactive atoms present in the specimen during the Mohenjodaro period.
Therefore, we can relate the decay constant, Aand time. t as:
$\frac{\text{N}}{\text{N}'}=\frac{\text{R}}{\text{R}'}=\text{e}^{-\lambda\text{t}}$
$\text{e}^{-\lambda\text{t}}=\frac{9}{15}=\frac{3}{5}$
$-\lambda\text{t}=\log_\text{e}\frac{3}{5}=-0.5108$
$\therefore\ \text{t}=\frac{0.5108}{\lambda}$
But $\lambda=\frac{0.693}{\text{T}_{1/2}}=\frac{0.693}{5730}$
$\therefore\ \text{t}=\frac{0.5108}{\frac{0.693}{5730}}=4223.5\text{ Years}$
Hence, the approximate age of the Indus-Valley civilisation is 4223.5 years.
View full question & answer→Question 23 Marks
The radionuclide $^{11}C$ decays according to
$^{11}_{6}\text{C}\rightarrow^{11}_{5}\text{B}+\text{e}^{+}+\text{v}:\ \text{T}_{1/2}=20.3 \text{ min}$
The maximum energy of the emitted positron is $0.960 MeV$.
Given the mass values:
$\text{m}(^{11}_{6})=10=11.011434\text{u}$ and $\text{m}(^{11}_{6}\text{B})=11.009305\text {u}.$
calculate $Q$ and compare it with the maximum energy of the positron emitted.
AnswerFor the given reaction, mass defect is,
$\Delta\text{m}=[\text{m}(^{11}_{6}-6\text{m}_\text{e})]-[\text{m}(^{11}_{5}\text{B})-5\text{ m}_\text{e}+\text{m}_\text{e}]$
$=\text{m}(^{11}_{6}\text{C})-\text{m}(^{11}_{5}\text{B})-2\text{m}_\text{e}$
$= 11.011434 u -11.009305 u - 2 \times 0.000548 u$
$= 0.001033 u$
Now, $Q-$ value is,
$Q = 0.001033 \times 931.5 MeV$
$= 0.962 MeV$
which, is the maximum energy of the positron.
We have,
$\text{Q}=\text{E}_\text{d}+\text{E}_\text{e}+\text{E}_\text{v}$
The daughter nucleus is too heavy compared to $e^+$ and $v$.
So, it carries negligible energy $(\text{E}_\text{d}\approx0).$ If the kinetic energy $(E_v)$ carried by the neutrino is minimum $($i.e., zero$),$ the positron carries maximum energy, and this is practically all energy $Q$.
Hence, maximum $\text{E}_\text{e}\approx\text{Q}.$
View full question & answer→Question 33 Marks
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are $^{24}_{12}\text{Mg }23.98504\text{u}),^{25}_{12}\text{Mg }(24.98584\text{u})\text{ and }^{26}_{12}\text{Mg }(25.98259\text{u}).$ The natural abundance of $^{24}_{12}\text{Mg}\text{ is }78.99\%$ by mass. Calculate the abundances of other two isotopes.
AnswerAverage atomic mass of magnesium, m = 24.312 u
mass of magnesium isotope $^{24}_{12}\text{Mg },\text{m}_1=23.98504\text{ u}$
mass of magnesium isotope $^{25}_{12}\text{Mg },\text{m}_2=24.98584\text{ u}$
mass of magnesium isotope $^{26}_{12}\text{Mg },\text{m}_3=25.98259\text{ u}$
Abundance of $^{24}_{12}\text{Mg },\eta_1=78.99\%$
Abundance of $^{25}_{12}\text{Mg },\eta_2=\text{x}\%$
Hence, abundance of $^{26}_{12}\text{Mg },\eta_3=100-\text{x}-78.99\%=(21.01-\text{x})\%$
We have the relation for the average atomic mass as:
$\text{m}=\frac{\text{m}_1\eta_1+\text{m}_2\eta_2+\text{m}_3\eta_3}{\eta_1+\eta_2+\eta_3}$
$24.312=\frac{23.98504\times78.99+24.98584\times\text{x}+25.98259\times(21.01=\text{ x})}{100}$
2431.2 = 1894.5783096 + 24.98584x + 545.8942159 = 25.98259x
0.99675x = 9.2725255
And $\therefore\ \text{x}\approx9.3\%$
Hence, the abundance of $^{25}_{12}\text{Mg}$ is 9.3% and that of $^{26}_{12}\text{Mg}$ is 11.71%.
View full question & answer→Question 43 Marks
$A 1000 MW$ fission reactor consumes half of its fuel in $5.00 y$. How much $^{235}_{92}\text{U}$ did it contain initially? Assume that the reactor operates $80\%$ of the time, that all the energy generated arises from the fission of $^{235}_{92}\text{U}$ and that this nuclide is consumed only by the fission process.
AnswerGiven,
Power of reactor $= 1000 MW = 10^3 MW$
$= 10^9 W$
$= 10^9 Js^{-1}$
Energy generated by reactor in $5$ Years $= 5 \times 365 \times 24 \times 60 \times 60 \times 10^9 J$
Average energy generated $= 200 MeV$
$= 200 \times 1.6 \times 10^{-13} J$
Number of fission taking place or number of $\text{U}^{235}$ nuclei required,
$=\frac{5\times365\times24\times60\times60\times10^9}{200\times1.6\times10^{-13}}$
$= 8.2125 \times 10^{26} \times 6$
$= 49.275 \times 10^{26}$
Mass of $6.023 \times 10^{23}$ nuclei of $U = 235 gm = 235 \times 10^{-3} \ kg$
Mass of $8.2125 \times 10^{26}$ nuclei of $U,$
$=\frac{235\times10^{-3}}{6.023\times10^{23}}\times6\times8.2125\times10^{26}$
$= 1932 \ kg$
$\frac{1}{2}$ of fuel $= 1932 \ kg$
Tatal fuel $= 3864 \ kg$.
View full question & answer→Question 53 Marks
From the relation $\text{R}=\text{R}_0\text{A}^{1/3}$ where $R_0$ is a constant and $A$ is the mass number of a nucleus, show that the nuclear matter density is nearly constant $($i.e. independent of $A).$
Answerwe have the expression for nuclear radius as:
$\text{R}=\text{R}_0\text{A}^{1/3}$
Where,
$R_0 =$ Constant.
$A =$ Mass number of the nucleus
Nuclear matter density, $\rho=\frac{\text{Mass of the nucleus}}{\text{Volume of the nucleus}}$
Let m be the average mass of the nucleus.
Hence, mass of the nucleus $= mA$
$\therefore\ \rho\frac{\text{mA}}{\frac{4}{3}\pi\text{R}^3}=\frac{3\text{mA}}{4\pi\Big(\text{R}_0\text{A}^{\frac{1}{3}}\Big)^3}=\frac{3\text{mA}}{4\pi\text{R}^3_0\text{A}}=\frac{3\text{m}}{4\pi\text{R}^3_0}$
Hence, the nuclear matter density is Independent of $A.$ It is nearly constant.
View full question & answer→Question 63 Marks
The nucleus $^{23}_{10}\text{Ne}$ decays by $\beta$ emission. Write down the $\beta-$decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
$\text{m}(^{23}_{10}\text{Ne})=22.994466\text{ u}.$
$\text{m}(^{23}_{10}\text{Na})=22.089770\text{ u}.$
AnswerThe $\beta$-decay of $^{23}_{10}\text{Ne}$ may be represented as:
$^{23}_{10}\text{Ne}\rightarrow\ ^{23}_{11}\text{Na }\ -\ ^{0}_{-1}\text{e }\ +\ \overline{\text{v}}\ +\ \text{Q}$
Ignoring the rest mass of antineutrino $(\overline{\text{v}})$ and electron, we get Mass defect,
$\Delta\text{m}=\text{m}(^{23}_{10}\text{Ne})-\text{m}(^{23}_{11}\text{Na})$
$= 22.994466 - 22.989770$
$= 0.004696 u$
$\therefore\ \text{Q}=0.004696\times931\text{ MeV}=4.372\text{ MeV}.$
This energy of $4.3792 MeV,$ is shared by $e^-$ and $\overline{\text{v}}$ pair because, $^{23}_{11}\text{Na}$ is very massive.
The maximum $K.E$. of $e^- = 4.372 MeV,$ when energy carried by $\overline{\text{v}}$ is zero.
View full question & answer→Question 73 Marks
Obtain approximately the ratio of the nuclear radii of the gold isotope $^{197}_{79}\text{Au}$ and the silver isotope $^{107}_{47}\text{Ag}.$
AnswerUsing the relation between the radius of nucleus and atomic mass,
$\text{R}\approx\text{ A}^{1/3}$
Atomic mass of gold,$ A_1 = 197$
Atomic mass of silver, $A_2 = 107$
$\therefore\ \frac{\text{R}_1}{\text{R}2}=\Big(\frac{\text{A}_1}{\text{A}_2}\Big)^{1/3}$
$=\Big(\frac{197}{107}\Big)^{1/3}=\big(1.84\big)^{1/3}$
Now, taking $\log$ on both sides
$\Rightarrow\ \log_{10}\Big(\frac{\text{R}_1}{\text{R}_2}\Big)=\log_{10}\big(1.84\big)^{1/3}$
$\Rightarrow\ \log_{10}\Big(\frac{\text{R}_1}{\text{R}_2}\Big)=\frac{1}{3}\log_{10}\big(1.84\big)$
$=\frac{1}{3}\times0.2648$
$=0.08827$
$\Rightarrow\ \frac{\text{R}_1}{\text{R}_2}=\text{ antilog}\ (0.08827)$
$= 1.23,$ which is the required ratio of the nucleii.
View full question & answer→Question 83 Marks
Write nuclear reaction equations for
- $\alpha\text{-decay of }\ ^{226}_{88}\text{Ra}$
- $\alpha-\text{decay of }\ ^{226}_{94}\text{Pu}$
- $\beta\text{-decay of }\ ^{32}_{15}\text{P}$
- $\beta\text{-decay of }\ ^{210}_{83}\text{Bi}$
- $\beta\text{-decay of }\ ^{11}_{6}\text{C}$
- $\beta^+\text{-decay of }\ ^{97}_{43}\text{Tc}$
- $\text{Electron capture of }\ ^{120}_{54}\text{Xe}$
Answera is a nucleus of helium $(_2\text{He}^4)$ and $\beta$ is an electron $(\text{e}^{-}\text{for}\beta\text{ and }\text{e}^{+}\text{for}\beta^{+}).$ In every adecay,there is a loss of 2 protons and 4 neutrons. $\beta^{+}$ -decav, there is a loss of 1 proton and a neutrino isemitted from the nucleus. In every $\beta^{-}$ -decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.
For the given cases, the various nuclear reactions can be written as:
- $_{88}\text{Ra}^{226}\xrightarrow{\ \ \ \ \ \ \ } \ _{86}\text{Rn}^{222}+_2\text{He}^4$
- $^{242}_{94}\xrightarrow{\ \ \ \ \ \ \ } \ ^{86}_{92}\text{U }+\ ^4_2\text{He}$
- $^{32}_{15}\text{P}\xrightarrow{\ \ \ \ \ \ \ } \ ^{32}_{16}\text{S }+\ \text{e}^{-}+ _\text{V}^{-}$
- $^{210}_{83}\text{B}\xrightarrow{\ \ \ \ \ \ \ } \ ^{210}_{84}\text{PO }+\ \text{e}^{-}+\ _\text{V}^{-}$
- $^{11}_{6}\text{C}\xrightarrow{\ \ \ \ \ \ \ } \ ^{11}_{5}\text{B }+\ \text{e}^{+}+\ \text{V}$
- $^{97}_{43}\text{Tc}\xrightarrow{\ \ \ \ \ \ \ } \ ^{97}_{42}\text{MO }+\ \text{e}^{+}+ \text{V}$
- $^{120}_{54}\text{Xe}+\text{e}^{+}\xrightarrow{\ \ \ \ \ \ \ } \ ^{120}_{53}\text{I }+\ \text{V}$
View full question & answer→Question 93 Marks
Calculate the height of the potential barrier for a head on collision of two deuterons. $($Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius $2.0 fm.$)$
AnswerSuppose, the two particles are fired at each other with the same kinetic energy $K.E,$
so that they are brought to rest by their mutual Coulomb repulsion when they are just touching each other.
Distance between the centres of two deutrons, during head on collision $, r = 2 \times $ radius
Therefore,
$r = 4 fm = 4 \times 10^{-15} m$
Charge on each deutron $, e = 1.6 \times 10^{-19} C$
Now, potential energy $=\frac{\text{e}^2}{4\pi\varepsilon_0\text{r}}$
$=\frac{9\times10^9(1.6\times10^{-19})^2}{4\times10^{-15}}$
$=\frac{9\times1.6\times1.6\times10^{-14}}{4\times1.6\times10^{-16}}\text{ keV}$
$= 360 keV$
Now, since, potential energy is equal to twice the kinetic energy of deutron.
This implies,
$K.E $ of each deutron $= \frac{360}{2}=180\text{ keV}$
which is a measure of the height of the coulomb barrier.
View full question & answer→Question 103 Marks
Obtain the binding energy $($in $MeV)$ of a nitrogen nucleus $(^{14}_7\text{N}),$ given m $(^{14}_7\text{N}) = 14.010307 u$
AnswerAtomic mass of nitrogen $\big(_7\text{N}^{14}\big), m = 14.00307 u$
A nucleus of nitrogen $_7\text{N}^{14}$ contains $7$ protons and $7$ neutrons.
Hence, the mass defect of this nucleus, $\Delta\text{m}=7\text{m}_\text{H}+7\text{m}_\text{n}-\text{m}$
Where,
Mass of a proton $, m_H = 1.007825 u$
Mass of a neutron $, m_n = 1.008665 u$
$\therefore\ \Delta\text{m}=7\times1.007825+7\times1.008665-14.00307$
$= 7.054775 + 7.06055 - 14.00307$
$= 0.11236 u$
But $1 u = 931.5 MeV/c^2$
$\therefore\ \Delta\text{m}=0.11236\times931.5\text{Me/Vc}^2$
Hence, the binding energy of the nucleus is given as
$\text{E}_\text{b}=\Delta\text{Mc}^2$
Where,
$c =$ Speed of light
$\therefore\ \text{E}_\text{b}=0.11236\times931.5\Big(\frac{\text{MeV}}{\text{c}^2}\Big)\times\text{c}^2$
$= 104.66334 MeV$
Hence, the binding energy of a nitrogen nucleus is $104.66334 MeV$.
View full question & answer→Question 113 Marks
The fission properties of $^{239}_{94}\text{Pu}$ are very similar to those of $^{235}_{92}\text{U}.$ The average energy released per fission is $180 MeV$. How much energy, in $MeV,$ is released if all the atoms in $1 \ kg$ of pure $^{239}_{94}\text{Pu}$ undergo fission?
AnswerGiven,
Average amount of energy released per fission, $^{239}_{94}\text{Pu}=180=\text{MeV}$
Quantity of fissionable material $= 1 \ kg$
In $239 gm Pu,$ number of fissionable atom or nuclei $= 6.023 \times 10^{23}$
In $1 g$ of $Pu,$ number of fissionable atom or nuclei $=\frac{6.023\times10^{23}}{239}$
In $1000 gm$ of $Pu,$ number of fissionable atom or nuclei,
$=\frac{6.023\times10^{23}}{239}\times1000$
$=25.2\times10^{23}$
Therefore,
Total energy released in fission of $25.2 \times 10^{23 }Pu$ nucleus or in fission of $1 \ kg$ pure $Pu$ is,
$= 180 \times 25.2 \times 10^{23}$
$= 4536 \times 10^{23} MeV$
$= 4.5 \times 10^{26} MeV.$
View full question & answer→Question 123 Marks
Boron has two stable isotopes, $^{10}_5\text{B }\text{ and }\ ^{11}_5\text{B}.$ Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of $^{10}_5\text{B }\text{ and }\ ^{11}_5\text{B}.$
AnswerMass of boron isotope $^{10}_5\text{B },\ \text{m}_1=10.01294\text{ u}$
Mass of boron isotope $^{11}_5\text{B },\ \text{m}_2=11.00931\text{ u}$
Abundance of $^{10}_5\text{B },\ \text{n}_1=\text{x}\%$
Abundance of $^{11}_5\text{B },\ \text{n}_2=(100-\text{x})\%$
Atomic mass of boron, m = 10.811 u
The atomic mass of boron atom is given as:
$\text{m}=\frac{\text{m}_1\text{n}_1+\text{m}_2\text{n}_2}{\text{n}_1+\text{n}_2}$
$10.811=\frac{10.01294\times\text{ x }+11.00931\times(100-\text{ x})}{\text{ x }+100-\text{ x}}$
1081.11 = 10.01294x + 1100.9312 - 11.00931x
$\therefore\ \text{x}=\frac{19.821}{0.99637}=19.89\%$
And 100 - X = 80.11%
Hence, the abundance of $^{10}_5\text{B is }19.89\%\text{ and that of }^{11}_5\text{B is }80.11\%.$
View full question & answer→Question 133 Marks
How long can an electric lamp of $100W$ be kept glowing by fusion of $2.0 \ kg$ of deuterium? Take the fusion reaction as:
$^2_1\text{H}+^2_1\text{H}\rightarrow\ ^3_2\text{He}+\text{n}+3.27\text{ MeV}$
AnswerPower of the electric lamp $= 100 W$
When two nuclei of deuterium fuse together, energy released $= 3.2 MeV$
Number of deuterium atoms in $2 \ kg$ is,
$=\frac{6.023\times10^{23}}{2}\times2000$
$= 6.023\times10^{26}$
Energy released when $6.023 \times 10^{26}$ nuclei of deuterium fuse together,
$=\frac{3.2}{2}\times6.023\times10^{26}\text{ MeV}$
$=\frac{3.2}{2}\times6.023\times10^{26}\times1.6\times10^{-13}\text{ J}$
$=1.54\times10^{14}\text{J} $ or ${Ws}$
If the lamp glows for time t, then the electrical energy consumed by the lamp is $100 \ t,$
$\therefore\ 100\text{t}=1.54\times10^{14}$
$\text{t}=1.54\times10^{12}\text{ S}$
$=\frac{1.54\times10^{12}}{3.154\times10^{7}}$ Years
$=4.88\times10^4$ Years.
which is the life span of an electric lamp.
View full question & answer→Question 143 Marks
Find the Q-value and the kinetic energy of the emitted α-particle in the $\alpha $-decay of
- $^{226}_{88}\text{Ra}$ and
- $^{220}_{86}\text{Rn}.$
Given
$\text{m}(^{226}_{88}\text{Ra})=226.02540\text{ u}.$ $\text{m}(^{222}_{86}\text{Rn})=222.01750\text{ u}.$
$\text{m}(^{222}_{86}\text{Rn})=220.01137\text{ u}.$ $\text{m}(^{216}_{84}\text{Po})=216.00189\text{ u}.$ Answer
- The reaction invoved is,
$^{226}_{88}\text{Ra}\ \rightarrow^{222}_{86}\text{Rn }\ + \ ^{4}_{2}\text{He}$
The difference in mass between the original nucleus and the decay products = 226.02540 u - (222.01750 u + 4.00260 u)
= + 0.0053 u
$\therefore\ $Energy equivalent or Q-value = 0.0053 × 931.5 MeV
= 4.93695 MeV
= 4.94 MeV
The decay products would emerge with total kinetic energy 4.94 MeV.
Momentum is conserved. If the parent nucleus is at rest, the daughter and the a-particle have momenta of equal magnitude p but, in the opposite direction.
Kinetic energy, $\text{K}=\frac{\text{p}^2}{2\text{m}}.$
Since, p is the same for the two particles therefore the kinetic energy divides inversely as their masses.
The $\alpha$-particle gets $\frac{222}{222+4}$ of the total i.e., $\frac{222}{226}\times4.94\text{ MeV }\text{ or }4.85\text{ MeV}.$
- The difference in mass between the original nucleus and the decay products = 220.01137 u - (216.00189 u + 4.00260 u)
= 0.00688 u
$\therefore\ $Q-value or Energy equivalent = 0.00688 × 931.5 MeV
= 6.41 MeV
Energy of the alpha particle, $\text{E}_\alpha=\frac{216}{216+4}\times6.41\text{ MeV}=6.29\text{ MeV}.$ View full question & answer→Question 153 Marks
The $Q$ value of a nuclear reaction $A + b \rightarrow C + d$ is defined by $Q = [ m_A + m_b – m_C – m_d]c^2$ where the masses refer to the respective nuclei. Determine from the given data the $Q-$ value of the following reactions and state whether the reactions are exothermic or endothermic.
- $^1_1\text{H}+^3_1\text{H}\rightarrow^2_1\text{H}+^2_1\text{H}$
- $^{12}_6\text{C}+^{12}_6\text{C}\rightarrow^{20}_{10}\text{Ne}+^4_2\text{He}$
Atomic masses are given to be
$\text{m}(^2_1\text{H})=2.014102\text{ u}$
$\text{m}(^3_1\text{H})=3.016049\text{ u}$
$\text{m}(^{12}_6\text{C})=12.000000\text{ u}$
$\text{m}(^{20}_{10}\text{Ne})=19.992439\text{ u}$ Answer
- Considering the first reaction,
$^1_1\text{H}\ +\ ^3_1\text{H}\rightarrow\ ^2_1\text{H}\ +\ ^2_1\text{H}$
$Q-$ value is given by,
$\text{Q}=\Delta\text{m}\times931.5\text{ MeV}$
$=[\text{m}(^1_1\text{H})+\text{m}(^3_1\text{H}-2\text{m}(^2_1\text{H})]\times931\text{ MeV}$
$=[1.007825+3.016049-2\times2.014102]\times931\text{ MeV}$
$=-4.03\text{ MeV}$
Since, $Q-$ value is negative, this reaction is endothermic.
- The second reaction is,
$^{12}_6\text{C}\ +\ ^{12}_6\text{C}\rightarrow\ ^{20}_{10}\text{Ne}\ +\ ^4_2\text{He}$
$Q-$ value is given by,
$\text{Q}=\Delta\text{m}\times931\text{ MeV}$
$=[2\text{m}(^{12}_6\text{C})-\text{m}(^{20}_{10}\text{Ne})-\text{m}(^4_2\text{He})]\times931\text{ MeV}$
$=[24.000000-19.992439-4.002603]\times931\text{ MeV}$
$=+\ 4.61\text{ MeV}$
Since, the $Q-$ value is positive, the reaction is exothermic. View full question & answer→Question 163 Marks
- Write the basic nuclear process involved in the emission of$\beta^{+}$ in a symbolic form, by a radioactive nuclear.
- in the reactions given below:
- $^{11}_{6}\text{C}\rightarrow ^{\text{z}}_{\text{y}}\text{B} + \text{x} + \text{v}$
- $^{12}_{6}\text{C}\rightarrow ^{12}_{6}\text{C}\rightarrow^{20}_{\text{a}}\text{Ne} + ^{c}_{b}\text{He}$
Find the values of x,y, and z and a,b and c.Answer
- Basic nuclear reaction.
$\text{P}\rightarrow\text{n} + \text{e}^{+} + \text{v}$
-
- $\text{x} = \beta^{+}/^{o}_{1}\text{e} , \text{y} = 5 , \text{z} =11 $
- $\text{a} = 10 , \text{b}= 2 , \text{c} = 4.$
View full question & answer→Question 173 Marks
Draw a plot of potential energy of a pair of nucleons as a function of their separations. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces.
Answer
For r < OB, repulsive.
For r > OB force is attractive.
Nuclear forces are.
- Very strong.
- Charge independent.
- Show saturation.
- Spin-dependent.
View full question & answer→Question 183 Marks
- Write symbolically the $\beta^– $ decay process of $\frac{15}{32}P$.
- Derive an expression for the average life of a radionuclide. Give its relationship with the half $-$ life.
Answer$\beta^–$ decay process
$^{32}_{15}\text{P}\rightarrow^{32}_{16}\text{S} + \text{e}^{-} + \overline{\text{v}}$ or $^{32}_{15}\text{P}\rightarrow^{32}_{15}\text{X} + _{-1}\text{e}^{0} + \overline{\text{v}}$
Derivation of average life:
$\tau=\frac{\lambda\text{N}_{0}\int\limits_{0}^{\infty}\text{te}^{-\lambda\text{t}}\text{dt}}{\text{N}_{0}} = \lambda\int\limits_{0}^{\infty}\text{te}^{-\lambda\text{t}}\text{dt}$
$\Rightarrow\tau = 1/ \lambda$
Relation of average life with half life:
$\text{T}_{1/2} = \frac{\ell\text{n}2}{\lambda} = \tau\ell\text{n} 2$
View full question & answer→Question 193 Marks
Draw a plot showing the variation of binding energy per nucleon versus the mass number A. Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion.
AnswerPlot:
Explanation: A very heavy nucleus, say A = 240, has lower binding energy per nucleon compared to that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightly bound.
This implies energy would be released in the process of fission.
Consider two very light nuclei $({\text{A}}\leq 10) $ joining to form a heavier nucleus. The binding energy per nucleon of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion. View full question & answer→Question 203 Marks
Draw the graph to show variation of binding energy per nucleon with mass number of different atomic nuclei. Calculate binding energy/nucleon of $^{40}_{20}\text{Ca}$ nucleus. Given: mass of $^{40}_{20}\text{Ca} = 39.962589\text{u}$.
mass of proton $= 1.007825 u.$
mass of neutron $= 1.008665 u. $
and $1 u = 931 MeV/C^{2.}$
AnswerGraph:
Calculation of $\bigtriangleup\text{m}$ for $^{40}_{20}\text{Ca}$.
$m = (20\times 1.007825 +20 \times 1.008665)-39.962589 = 0.367211 amu$.
$BE = 0.367211\times 931Mev. = 341.87 Mev$.
$BE$ per nucleon $= 8.547Mev.$ View full question & answer→Question 213 Marks
A neutron is absorbed by a $^{6}_{3}\text{Li}$ nucleus with the subsequent emission of an alpha particle.
- Calculate the energy released, in $MeV,$ in this reaction.
- Write the corresponding nuclear reaction.
Given: mass $^{6}_{3}\text{Li} =6015126 u;$ mass $($neutron$) =1.0086654 u;$ mass $($alpha particle$) = 4.0026044 u$ and mass $($triton$) = 3.0100000 u.$
Take $1 u = 931\ MeV/C^2.$ Answer$^{6}_{3}\text{Li} + ^{1}_{0}\text{n}\rightarrow^{3}_{1}\text{He} + \text{Q}$
$\triangle m = ($mass of $^{6}_{3}\text{Li} +$ mass of neutrons$) \ – ($mass of $\alpha$ particle $+$ mass of $^{3}_{1}\text{H})$
$\triangle m = 0.011187u$
Energy released $\text{Q} =\bigtriangleup\text{m}\times931$
$= 10.415\ \text{MeV.}$
View full question & answer→Question 223 Marks
Define the terms half$-$life period and decay constant of a radioactive substance. Write their $S.I.$ units. Establish the relationship between the two.
AnswerHalf life$-$It is the time at which number/amount of radioactive nuclei/sample at any time reduces to one half its initial value.
Unit $-$ second
Decay constant – It is the ratio of the rate at which the number of atoms will decay to the total number of atoms present at that time. or It is the reciprocal of time in which the radioactive sample reduces to $\bigg(\frac{1}{\text{e}}\bigg)^{\text{th}}$of its intial value.
Unit $–$ second$^{-1}$ Derivation: $\text{N}\big(\text{t}\big) =\text{N}_\circ\text{e}^{-\lambda\text{t}}$
$\text{R} = - \frac{\text{dN}}{\text{dt}} =\lambda\text{N}_{\circ}\text{e}^{-\lambda\text{t}}..........................................(1) \text{N} =\frac{\text{N}_{\circ}}{2}$ at $\text{t} = \text{T}_{1/2}$
$\therefore \text{T}_{1/2} = \frac{\text{\log}_{e}2}{\lambda} = \frac{0.693}{\lambda}$.
View full question & answer→Question 233 Marks
- Derive the relation between the decay constant and half life of a radioactive substance.
- A radioactive element reduces to 25% of its initial mass in 1000 years. Find its half life.
Answer
- ${N}(t)=N_0\text{ }e^{-\lambda{t}}$
When $t=T_{1/2}\Rightarrow{N}(t)=\frac{N_0}{2}$
$\therefore {\text{ }} \frac{N_0}{2}=N_0\text{ }e^{-\lambda}T_{1/2}$
$\Rightarrow\frac{1}{2}=e^{-\lambda}T_{1/2}$
$\Rightarrow-{\lambda}T_\frac{1}{2}=-ln2$
$\Rightarrow\text{ }T_\frac{1}{2}=\frac{ln2}{\lambda}$
$\Rightarrow\frac{0.693}{\lambda}$
- $\frac{N}{N_0}=\bigg(\frac{1}{2}\bigg)^n\text{ }\text{ }\text{ }\text{ }n=\frac{t}{T_{1/2}}$
$\text{Given}\text{ }\frac{N}{N_0}=\frac{1}{4}=\bigg(\frac{1}{2}\bigg)^n $
$\bigg(\frac{1}{2}\bigg)^n=\bigg(\frac{1}{2}\bigg)^2$
$\therefore \text{Number of half lives}=2$
$\Rightarrow\frac{1000}{T_{1/2}}=2$
$\Rightarrow{T}_\frac{1}{2}=\frac{1000}{2}=500 \text{ }\text{years}$
[?????????????
1000 years = 2 half lives
$\therefore$ Half life = 500 years] View full question & answer→Question 243 Marks
- Write the process of $\beta$ decay. How can radioactive nuclei emit $\beta$-particles even though they do not contain them? Why do all electrons emitted during $\beta$-decay not have the same energy?
- A heavy nucleus splits into two lighter nuclei. Which one of the two - parent nucleus or the daughter nuclei has more binding energy per nucleon?
Answer
- A nucleus, that spontaneously decays by emitting an electron, or a positron, is said to undergo $\beta$ decay
[Alternatively $ ^{A}_{Z}\text{X}\longrightarrow^{\text{ }\text{ }\text{ }\text{ }\text{ }A}_{Z+1}\text{Y}+e^-+\bar{v}$
$ ^{A}_{Z}\text{X}\longrightarrow^{\text{ }\text{ }\text{ }\text{ }\text{ }A}_{Z-1}\text{Y}+e^++v$ (antineutrino)
During β decay, nucleons undergo transformation. We can have:
${n}\text{ }{\longrightarrow}\text{ }{p}+e^-+{\bar{v}}$
$ \longrightarrow$ A neutron converts into a proton and an electron [Alternatively
${p}\longrightarrow{n}+e^++v$
[A proton converts into a neutron and a positron] It is because the neutrinos, or antineutrino, carry off different amounts of energy.
- The daughter nuclei have more binding energy per nucleon.
View full question & answer→Question 253 Marks
A proton and an $\alpha$-particle move perpendicular to a magnetic field. Find the ratio of radii of the circular paths described by them when both (i) have equal momenta, and (ii) were accelerated through the same potential difference.
Answer
- $\frac{mv^{2}}{r}=???$.
$\therefore{r}=\frac{mv}{??}=\frac{?}{??}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(p = mv)$
For proton $r_p=\frac{p}{q_p{B}}$
For $\alpha$ particles ${r}_{\alpha}=\frac{p}{q_{\alpha}{B}}$
$\therefore\text{ }\frac{r_p}{r_{\alpha}}=\frac{q_{\alpha}}{q_p}=2$
- $r=\frac{??}{??}=\frac{1}{B}\sqrt{\frac{2??}{?}}$
for proton $r_p=\frac{1}{B}\sqrt\frac{2m_p{V}}{q_p}$
and for $\alpha$ particles $r_\alpha=\frac{1}{B}\sqrt\frac{2m_{\alpha}{V}}{q_{\alpha}}$
$ \therefore\frac{r_p}{r{\alpha}}=\sqrt{\frac{m_p}{q_p}\frac{q_{\alpha}}{m{\alpha}}}$
$\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}$ View full question & answer→Question 263 Marks
- Define ‘activity’ of a radioactive substance.
- Two different radioactive elements with half lives $T_1$ and $T_2$ have $N_1$ and $N_2$ undecayed atoms respectively present at a given instant. Derive an expression for the ratio of their activities at this instant in terms of $N_1$ and $N_2.$
Answer
- Number of radioactive nuclei decaying per second at any time.
- $R_1=\lambda_1,N_1=\frac{0.693}{T_1}N_1$
$R_2=\lambda_2,N_2=\frac{0.693}{T_2}N_2$
$\frac{R_1}{R_2}=\frac{N_1}{N_2}\times\frac{T_2}{T_1}$ View full question & answer→Question 273 Marks
- State the law of radioactive decay. Write the SI unit of ‘activity’.
- There are $4\sqrt{2}\times10^6$ radioactive nuclei in a given radioactive sample. If the half life of the sample is 20 s, how many nuclei will decay in 10 s?
Answer
- Statement: Rate of decay of a given radioactive sample is directly propotional to the total number of undecayed nuclei present in the sample.
- $N=N_0e^{-\lambda t}/\frac{N}{N_0}=\Big(\frac{1}{2}\Big)^n$
$\text{n}=\frac{t}{T_{1/2}}=\frac{10}{20}=\frac{1}{2}$
$\Rightarrow N=4\sqrt{2}\times10^{6}\times\Big(\frac{1}{2}\Big)^{\frac{1}{2}}$
$=4\times10^6$ nuclei. View full question & answer→Question 283 Marks
- A radioactive nucleus $'A\ '$ undergoes a series of decays as given below:
- $\text{A}\xrightarrow\alpha\text{A}_1\xrightarrow\beta\text{A}_2\xrightarrow\alpha\text{A}_3\xrightarrow\gamma\text{A}_4$
The mass number and atomic number of $A_2$ are $176$ and $71$ respectively. Determine the mass and atomic numbers of $A_4$ and $A$.
- Write the basic nuclear process underlying $\beta+$ and $\beta-$ decays.
Answer$\text{A}^{180}_{74}\xrightarrow\alpha\text{A}^{176}_{72}\xrightarrow\beta\text{A}^{176}_{71}\xrightarrow\alpha\text{A}^{172}_{69}\xrightarrow\gamma\text{A}^{172}_{69}$
The mass number and atomic number of $A_4$ is $172$ and $69,$ respectively.
The mass number and atomic number of $A$ is $180$ and $74,$ respectively.
Basic process underlying $\beta+$ and $\beta-$ decay are
During a weak interaction an atomic nucleus converts into a nucleus with one higher atomic number while emitting one electron and an electron antineutrino this is called beta minus decay.
$X^A_Z\rightarrow Y^A_{Z+1}+e^-+\bar{ve}$
During a weak interaction an atomic nucleus converts into a nucleus with one lower number while emitting a positron and electron neutrino this is called beta Plus decay.
$X^A_Z\rightarrow Y^A_{Z-1}+e^++\bar{ve}$
View full question & answer→Question 293 Marks
Answer the following:
- Name the em waves which are produced during radioactive decay of a nucleus. Write their frequency range.
- Welders wear special glass goggles while working. Why? Explain.
- Why are infrared waves often called as heat waves? Give their one application.
Answer
- $\gamma-rays$
Range: $10^{19}$ to $10^{23} Hz$
- To protect the eyes from large amount of $UV$ radiations produced by welding arcs.
- Because water molecules present in the materials readily absorb the infra red rays get heated up.
View full question & answer→Question 303 Marks
- Deduce the expression,$ N = N_0 \text{e}^{-\lambda\text{t}},$ for the law of radioactive decay.
-
- Write symbolically the process expressing the $B^+$ decay of $^{22}_{11}\text{Na},$ Also write the basic nuclear process underlying this decay.
- Is the nucleus formed in the decay of the nucleus $^{22}_{11}\text{Na},$ isotope or isobar?
Answer
-
$\frac{\text{dN}}{\text{dt}} = -\lambda\text{N}$
$\int_{N_{0}}^{N}\frac{\text{dN}}{\text{N}} = \int_{0}^{t} - \lambda\text{dt}$
$[\log_{e}\text{N}]_{N_{0}}^{N} = - \lambda[\text{t}]_{0}^{t}$
$\log_{e}\frac{\text{N}}{\text{N}_{0}} = - \lambda\text{t}$
$\text{N} = \text{N}_{0}\text{e}^{-\lambda\text{t}}$
-
- $^{22}_{11}\text{Na}\rightarrow^{22}_{11}\text{Ne} + \text{e}^{+} + \text{v}$
Also accept, if a student does not identify the product nucleus and writes as
$^{22}_{11}\text{Na}\rightarrow^{22}_{10}\text{X} + \text{e}^{+} + \text{v}$
Basic process
$\text{p}\rightarrow\text{n} + \text{e}^{+} +\text{v}$
- Isobar.
View full question & answer→Question 313 Marks
- In a typical nuclear reaction, e.g.
- $^{2}_{1}\text{H} +^{2}_{1}\text{H}\rightarrow^{3}_{2}\text{He} + \text{n} + 3.27\text{MeV}$
although number of nucleons is conserved, yet energy is released. How? Explain.
- Show that nuclear density in a given nucleus is independent of mass number $A.$
Answer
- In nuclear reaction
$^{2}_{1}\text{H} +^{2}_{1}\text{H}$
$\rightarrow^{3}_{2}\text{He} + \text{n} + 3.27\text{MeV}$
Cause of the energy released:
- Binding energy per nucleon of ${3}_{2}\text{ He}$ becomes more than the $(BE/A)$ of
- ${2}_{1}\text{ H}.$
- Mass defect between the reactant and product nuclei.
$\Delta\text{E} = \Delta\text{mC}^{2}$
$ = \big[2\text{m}(^{2}_{1}\text{H}) - \text{m}(^{3}_{2}\text{He}) + \text{m}(\text{n})\big]\text{C}^{2}$
- The radius of nucleus of mass number $A$ is given by $R = R_0A^{1/3}$
Volume of the nucleus $\text{V} = \frac{4}{3}\pi\text{R}^{3} = \frac{4}{3}\pi\text{R}_{0}^{3}\text{A}$
Density of the matter in the nucleus
$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{\text{A}{(u)}}{\frac{4}{3}\pi\text{R}_{0}^{3}\text{A}}$
$\rho = \frac{1}{\frac{4}{3}\pi\text{R}_{0}^{3}} = \frac{3}{4\pi\text{R}_{0}^{3}}$
The expression of the density is independent of mass number $A$. View full question & answer→Question 323 Marks
- What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number 'A' lying 30
- Show that the density of nucleus over a wide range of nuclei is constantindependent of mass numberA.
Answer
- Saturation/short range nature of nuclear forces.
- We have
$\text{R} = \text{R}_{0}\text{A}^{1/3}$
$\therefore\text{ Density} \rho=\frac{\text{mA}}{\frac{4}{3}\pi\big(\text{R}_{0}\text{A}^{1/3}\big)^{3}}$
$ =\frac{\text{m}}{\frac{4}{3}\pi\text{R}_{0}^{3}}$
Hence $\rho$ is independent of A.
(Here m is the mass of the nucleus.) View full question & answer→Question 333 Marks
State the law of radioactive decay.
Plot a graph showing the number $(N)$ of undecayed nuclei as a function of time $(t)$ for a given radioactive sample having half life $T_\frac12.$
Depict in the plot the number of undecayed nuclei at $(i) t = 3 T_\frac12$ and $(ii) t = 5 T_\frac12.$
AnswerThe number of nuclei undergoing decay per unit time, at any instant, is proportional to the total number of nuclei in the sample at that instant.
Alternate Answer
$-\frac{\text{dN}}{\text{dt}}\alpha\text{N}$
$\Rightarrow\frac{\text{dN}}{\text{dt}} = - \lambda\text{N}$
View full question & answer→Question 343 Marks
- Define ‘activity’ of a radioactive material and write its $S.I.$ unit.
- Plot a graph showing variation of activity of a given radioactive sample with time.
- The sequence of stepwise decay of a radioactive nucleus is
If the atomic number and mass number of $D_2$ are $71$ and $176$ respectively, what are their corresponding values for $D$?Answer
- The total decay rate $($of a sample$)$ at the given instant, i.e., the number of radionuclides disintegrating per unit time is called the activity of that sample. The $SI$ unit for activity is becquerel $(Bq)$.
- Graph:
- $72$ and $180$.
View full question & answer→Question 353 Marks
Calculate the amount of energy released during the $\alpha - $ decay of $^{238}_{92}\text{U}\to ^{234}_{92}\text{Th} + ^{4} _{2}\text{He}$Given:
- Atomic mass of $^{238}_{92}\text{U}= 238.05079u$
- Atomic mass of $^{234}_{90}\text{Th}= 234.04363u$
- Atomic mass of $^{4}_{2}\text{He}= 4.00260\text{ u}$
$\text{1u} = 931.5 \text{MeV/c}^{2}$
Is this decay spontaneous? Give reason. Answer$\bigtriangleup\text{m} = \text{M}_{U} - \text{M}_{Th} - \text{M}_{He}$= 0.00456 u
Enetgy released $=\bigtriangleup\text{mc}^{2}$
= 0.00456 X 931.5
= 4.25 MeV
Yes
As $\bigtriangleup\text{m}$ is positive.
View full question & answer→Question 363 Marks
Explain, with the help of a nuclear reaction in each of the following cases, how the neutron to proton ratio changes during (i) alpha-decay (ii) beta-decay?
AnswerFor $\alpha$ decay
$^{238}_{92}\text{U}\to^{234}_{90}\text{Th}+ ^{4}_{2}\text{He}+\text{Q}$
(or any other nuclear reaction)
$\text{For} ^{238}_{92}\text{U},\frac{\text{n}}{\text{p}} = \frac { 238 - 92}{92} = \frac{146}{92} =1.587$
$\text{For} ^{234}_{90}\text{Th},\frac{\text{n}}{\text{p}} = \frac { 234 - 90}{90} = 1.6$
$\therefore \frac{\text{n}}{\text{p}} $ increases during $\alpha$ decay.
For $\beta$ decay
$^{60}_{27}\text{Co}\to^{60}_{28}\text{Ni}+ ^{0}_{-1}\text{e}+ \text{Q}$
(or any other nuclear reaction)
$\text{For} ^{60}_{27}\text{Co},\frac{\text{n}}{\text{p}} = \frac{60 - 27 }{27} = 1.22$
$\text{For} ^{60}_{28}\text{Ni},\frac{\text{n}}{\text{p}} = \frac{60 - 28 }{28} = 1.14$
$\therefore \frac{\text{n}}{\text{p}} $ decreases during $\beta$ decay
Alternate Answer
Full credit to be given if explained in terms of.
$^{\text{A}}_\text{Z}\text{X}\to^{\text{A-4}}_\text{Z-2}\text{Y}+ ^{4}_{2}\text{He}$
and
$^{\text{A}}_\text{Z}\text{X}\to^{\text{A}}_\text{Z+1}\text{Y}+ ^\circ_{1}\text{e} + {\overline{\text{V}}}\big(\text{V}\big)$
$\text{for }\alpha\text{ decay and }\beta\text{ decay}.$
View full question & answer→Question 373 Marks
Why is the mass of a nucleus always less than the sum of the masses of its constituents, neutrons and protons?If the total number of neutrons and protons in a nuclear reaction is conserved, how then is the energy absorbed or evolved in the reaction? Explain.
Answer
- The strong attractive nuclear forces act to bring the nucleons together to form the nucleus. This work is done at the expense of some mass of the (free) nucleons getting converted into energy. This results in a decrease of mass.
Alternate Answer
The binding energy needed to hold the nucleons together to form the nucleus, is obtained at the expense of some mass of the (free) nucleons getting converted into energy.
Alternate Answer
The forces of attraction, holding the nucleons together to form the nucleus gives the system a negative potential energy which is due to the mass lost in the process of formation of the nucleus.
Alternate Answer
When the free nucleons are brought together to form the nucleus, some energy gets released. This energy is released at the expense of the some mass of the free nucleons getting converted into energy.
- Even when the total number of neutrons and protons is conserved in a nuclear reaction the sum total of the masses of the products is different (either less or more) than the masses of the reactants. It is the energy equivalent of the mass difference (loss or gain) that gets released or absorbed in the reaction.
View full question & answer→Question 383 Marks
Draw a graph showing the variation of binding energy per nucleon with mass number for different nuclei. Explain, with the help of this graph, the release of energy by the process of nuclear fusion.
Answer
In case of nuclei of low atomic number, binding energy per nucleon is quite small; when fuse together, they form a nucleus of higher atomic mass and higher B.E./Nucleon, and hence, they release energy and become more stable.
View full question & answer→Question 393 Marks
- Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/ A) versus the mass number A.
- A radioactive isotope has a half-life of 10 years. How long will it take for the activity to reduce to 3·125%?
Answer
-
Nuclear fission: Binding energy per nucleon is smaller for heavier nuclei (greater than 200) i.e., heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, they become stable, this happens in nuclear fission.
Nuclear fusion: The binding energy per nucleon is small for light nuclei (between 2 to 20), i.e., they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher binding energy per nucleon of the latter results in the release of energy. This is what happens in a nuclear fusion.
- $\frac{\text{R}}{\text{R}_0}=\frac{\text{N}}{\text{N}_0}=\Big(\frac{1}{2}\Big)^{\text{n}}$
$\frac{\text{R}}{\text{R}_0}=3.125\%=\frac{3.125}{100}=\frac{1}{32}=\Big(\frac{1}{2}\Big)^5$
$=\Big(\frac{1}{2}\Big)^\text{n}=\Big(\frac{1}{2}\Big)^5\Rightarrow\ \text{n}=5$
$\text{t}=\text{nT}_{\frac{1}{2}}=5\times10=50\text{ year}$ View full question & answer→Question 403 Marks
The nucleus $^{235}_{92}\text{Y},$ initially at rest, decays into $^{231}_{90}\text{X}$ by emitting an $\alpha-$particle
$^{235}_{92}\text{Y}\xrightarrow{ \ \ \ \ \ \ \ \ } \ ^{231}_{90}\text{X}+ \ ^4_2\text{He}+\text{energy}.$
The binding energies per nucleon of the parent nucleus, the daughter nucleus and $\alpha-$particle are $7.8\ MeV, 7.835\ MeV$ and $7.07\ MeV,$ respectively. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted $\alpha-$particle. $($Mass of $\alpha-$ particle $= 6.68 \times 10^{–27}kg).$
AnswerThe binding energies per nucleon of the parent nucleus, the daughter nucleus and $\alpha-$particle are $7.8\ MeV, 7.835\ MeV$ and $7.07\ MeV,$ respectively.
Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted $\alpha-$particle. $($Mass of $\alpha-$particle $6.68 \times 10^{-27}kg),$
Energy released $= Q = 7.835 \times 231 + 7.07 \times 4 - 7.8 \times 235$
$\Rightarrow Q = 1809.885 + 28.28 - 1833$
$= 5.165\ MeV$
$= 5.165 \times 1.6 \times 10^{-13}J$
This energy will be taken away by $\alpha-$particle as kinetic energy.
$\therefore\frac{1}{2}\text{mv}^2=\text{Q}$
$\Rightarrow$ Speed of $\alpha-$particle,
$\text{v}=\sqrt{\frac{5.165\times1.6\times10^{-13}\times2}{6.68\times10^{-27}}}$
$=\sqrt{\frac{16.528}{6.68}\times10^{14}}$
$=\sqrt{2.474}\times10^7$
$=1.573\times10^7\text{m/ s}.$
View full question & answer→Question 413 Marks
Define the term ‘decay constant’ of a radioactive sample. The rate of disintegration of a given radioactive nucleus is $10000$ disintegrations/ s and $5,000$ disintegrations/ s after $20hr$. and $30hr$. respectively from start. Calculate the half life and initial number of nuclei at $t = 0.$
AnswerDecay constant is the fraction of the number of atoms that decay in one second. It is denoted by $\lambda.$
Let $N_0$ be the initial number of nuclei,
Let $\lambda$ be the decay constant
Let $\text{t}_\frac{1}{2}$ be the half $-$ life.
The instantaneous activity of radioactive material is given by:
$\text{A}=\text{A}_0\text{e}^{-\lambda\text{t}}$
Where $, A_0$ is activity at $t = 0$
Here, Activity after $20$ hours is $10,000 $ disintegrations per second
$\Rightarrow10,000=\text{A}_0\text{e}^{-\lambda(20\times3600)}\ ...(1)$
Activity after $30$ hours is $5,000$ disintegrations per second
$\Rightarrow5,000=\text{A}_0\text{e}^{-\lambda(30\times3600)}\ ...(2)$
On dividing $(1)$ by $(2),$
$2=\text{e}^{-\lambda\times3600}$
$\Rightarrow\lambda=\frac{\text{In}\ 2}{36000}=1.92\times10^{-5}$
So half $-$ life $\frac{\text{In}\ 2}{1.92\times10^{-5}}=36,000\text{s}=10\ \text{hours}$
We know that, $\frac{\text{dN}}{\text{dt}}=\lambda\text{N}$
$\Rightarrow10,000=(1.92\times10^{-5})\times\text{N}_1$
$\Rightarrow\text{N}_1=\frac{10,000}{1.92\times10^{-5}}=5.208\times10^8$
As the half $-$ life is $10$ hours, thus the initial numbers of nuclei $, N_0 = 2N_1$
$\Rightarrow N_0 = 10.416 \times 10^8$
View full question & answer→Question 423 Marks
An ideal diode should pass a current freely in one direction and should stop it completely in the opposite direction. Which is closer to ideal-vacuum diode or a p-n junction diode?
AnswerIt should be an ideal vacuum diode. When a p-n junction diode is reverse biassed then a small current called reverse current flows across the diode. As the the p‒n junction diode allows some current in reverse biassed condition also so the given diode cannot be a p-n junction diode.
View full question & answer→Question 433 Marks
- A giant refracting telescope at an observatory has an objective lens of focal length $15\ m$. If an eyepiece of focal length $1.0 \ cm$ is used, what is the angular magnification of the telescope?
- If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^{6 }m,$ and the radius of lunar orbit is $3.8 \times 10^{8 }m.$
AnswerFocal length of the objective lens $, f_0 = 15 m = 15 \times 10^{2 }cm$ Focal length of the eyepiece $, f_e = 1.0 \ cm$
- The angular magnification of a telescope is given as:
$\alpha=\frac{\text{f}_0}{\text{f}_\text{e}}$
$=\frac{15\times10^2}{1.0}=1500$
Hence, the angular magnification of the given refracting telescope is $1500.$
- Diameter of the moon $, d = 3.48 \times 10^{6 }m$
Radius of the lunar orbit $, r_0 = 3.8 \times 10^8 m$
Let $d\ '$ be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
$\frac{\text{d}}{\text{r}_0}=\frac{\text{d}'}{\text{d}_0}$
$\frac{3.48\times10^6}{3.8\times10^8}=\frac{\text{d}'}{15}$
$\therefore \ \text{d}'=\frac{3.48}{3.8}\times10^{-2}\times15$
$= 13.74 \times 10^{-2} m = 13.74 \ cm$
Hence, the diameter of the moon's image formed by the objective lens is $13.74 \ cm.$ View full question & answer→Question 443 Marks
The following table shows some measurements of the decay rate of a radionuclide sample. Find the disintegration constant.
| Time (min) |
lnR (Bq) |
| 36 |
5.08 |
| 100 |
3.29 |
| 164 |
1.54 |
| 218 |
0 |
Answer$\text{R}=\text{R}_0\text{e}^{-\lambda\text{t}}$
In $\text{R}=1\text{n}\text{ R}_0^{-\lambda\text{t}}$
In $\text{R}=-\lambda \text{t}+1\text{n}\text{R}_0$
Slope of In $\text{R}\frac{\text{v}}{\text{s}}\text{t}\text{ is } -\lambda'$
$-\lambda =\frac{0-1.52}{218-164}$
$\Rightarrow \lambda= 0.02\text{minute}^{-1}$
View full question & answer→Question 453 Marks
The half$-$life of $^{14}_6\text{C}$ is $5700$ years. What does it mean? Two radioactive nuclei $X$ and $Y$ initially contain an equal number of atoms. Their half$-$lives are $1$ hour and $2$ hours respectively. Calculate the ratio of their rates of disintegration after two hours.
AnswerThe half$-$life of $^{14}_6\text{C}$ is $5700$ years.
It means that one half of the present number of radioactive nuclei of $^{14}_6\text{C}$ will remain undecayed after $5700$ years.
Number of nuclei $X$ after $2$ hours$, N_X =\text{N}_0\Big(\frac{1}{2}\Big)^{\frac{1}{\frac{\text{T}}{2}}}=\frac{\text{N}_0}{4}$
Number of nuclei $Y$ after $2$ hours$, N_Y =\text{N}_0\Big(\frac{1}{2}\Big)^{\frac{2}{2}}=\frac{\text{N}_0}{2}$
$\therefore$ Ratio of rates of disintegration $\frac{\text{R}_\text{X}}{\text{R}_\text{Y}}=\frac{\frac{\text{N}_0}{4}}{\frac{\text{N}_0}{2}}=\frac{1}{2}$
View full question & answer→Question 463 Marks
If three helium nuclei combine to form a carbon nucleus, energy is liberated. Why can't helium nuclei combine on their own and minimise the energy?
AnswerWhen three helium nuclei combine to form a carbon nucleus, energy is liberated. This energy is greater than the that liberated when these nuclei combine on their own. Hence, formation of carbon nucleus leads to much more stability as compared to the combination of three helium nuclei.
View full question & answer→Question 473 Marks
Calculate the binding energy per $^{40}_{20}\text{CA}$ nucleon nucleus.
$[$Given: $m \big(^{40}_{20}\text{Ca}\big)=39.962589\text{u}]$
$m_n($mass od a neutron$) = 1.008665u$
$m_p($mass of a proton$) =1.007825u$
$1u = 931\ MeV/c^2]$
AnswerTotal Binding energy of $^{40}_{20}\text{Ca}$ nucleus $=20\text{m}_\text{p}+20\text{m}_\text{n}-\text{M}\big(^{40}_{20}\text{Ca}\big)$
$= 20 \times 1.007825 + 20 \times 1.008665 - 39.962589$
$= 0.367211 u = 0.367211 \times 931\ MeV = 341.87\ MeV$
$\therefore$ Binding energy per nucleus $=\frac{341.87}{40}\text{MeV}/ \text{nucleon}$
$=8.55\ \text{MeV}/$ nucleon
View full question & answer→Question 483 Marks
What is obtained by fusion of two deuterons?
AnswerBy fusion of two deuterons, either tritium $\big(^3_1\text{H}\big)$ or an isotope of helium $\big(^3_1\text{He}\big)$ is obtained with release of energy. The reactions are:
$^2_1\text{H}\ +\ ^2_1\text{H}\ \rightarrow\ ^3_1\text{H}\ +\ ^1_1\text{H}\ +\ 4.03\text{ MeV}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(tritium)}$
$^2_1\text{H}\ +\ ^2_1\text{H}\ \rightarrow\ ^3_1\text{He}+^1_0\text{n}\ +\ 3.27\text{ MeV}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(isotope of helium)}$
View full question & answer→Question 493 Marks
Calculate the energy released if $U^{238}-$nucleus emits an $\alpha-$particle. or
Calculate the energy released in $MeV$ in the following nuclear reaction. $^{238}_{92}\text{U}\ \rightarrow\ ^{234}_{90}\text{Th}\ +\ ^4_2\text{He}\ +\ \text{Q}$ Given Atomic mass of $^{238}U = 238.05079u$ Atomic mass of $^{234}Th = 234.04363u$ Atomic mass of alpha particle $= 4.00260u 1u = 931.5\ MeV/ c^2$ Is the decay spontaneous? Give reason.
AnswerThe process is $^{238}_{92}\text{U}\ \rightarrow \ ^{234}_{90}\text{Th}\ +\ ^4_2\text{He}+\text{Q}$
The energy released $(\alpha-$particle$)$
$\text{Q}=(\text{M}_\text{U}-\text{M}_{\text{TH}}-\text{M}_{\text{He}})\text{c}^2$
$=(238.05079-234.0463-4.00260)\text{u}\times\text{c}^2$
$=(0.00456\text{u})\times \text{c}^2$
$=0.00456\times\Big(\frac{931.5\text{MeV}}{\text{c}}^2\Big)\text{c}^2$
$=4.25\ \text{MeV}$
Yes, the decay is spontaneous $($since $Q$ is positive$)$.
View full question & answer→Question 503 Marks
Calculate the energy released by $1g$ of natural uranium assuming $200\ MeV$ is released in each fission event and that the fissionable isotope $^{235}U$ has an abundance of $0.7\%$ by weight in natural uranium.
Answer$1g$ of $'I\ ’$ contain $0.007g\ \ U^{235}$
So, $235g$ contains $6.023 \times 1023$ atoms.
So, $0.7g$ contains $\frac{6.023\times10^{23}}{235}\times0.007\text{ atom}$
$1$ atom given $200Mev.$
So, $0.7g$ contains $\frac{6.023\times10^{23}\times0.007\times200\times10^6\times1.6\times10^{-19}}{235}\text{J}=5.74\times10^{-8}\text{J}$
View full question & answer→Question 513 Marks
A uranium reactor develops thermal energy at a rate of $300\ MW$. Calculate the amount of $^{235}U$ being consumed every second. Average released per fission is $200\ MeV$.
AnswerLet $n$ atoms disintegrate per second
Total energy emitted$/sec = (n \times 200 \times 106 \times 1.6 \times 10^{-19})J =$ Power
$300\ MW = 300 \times 106$ Watt $=$ Power
$300 \times 106 = n \times 200 \times 106 \times 1.6 \times 10^{-19}$
$\Rightarrow\text{n}=\frac{3}{2\times1.6}\times10^{19}=\frac{3}{3.2}\times10^{-19}$
$6 \times 10^{23}$ atoms are present in $238$ grams
$\frac{3}{3.2}\times10^{19}$ atoms are present in $\frac{238\times3\times10^{19}}{6\times10^{23}\times3.2}=3.7\times10^{-4}\text{g}=3.7\text{mg}.$
View full question & answer→Question 523 Marks
The half-life of a radioisotope is 10h. Find the total number of disintegrations in the tenth hour measured from a time when the activity was 1Ci.
Answer$\text{t}_{\frac{1}{2}}=10\text{ hours, A}_0=1\text{ci}$
Activity after 9 hours $=\text{A}_0\text{e}^{-\lambda\text{t}}=1\times\text{e}^{\frac{-0.693}{10}\times9}=0.5359=0.539\text{ci}.$
No. of atoms left after 9th hour, $\text{A}_{9}=\lambda\text{N}_{9}$
$\Rightarrow\text{N}_9=\frac{\text{A}_9}{\lambda}=\frac{0.536\times10\times3.7\times10^{10}\times3600}{0.693}$
$=28.6176\times10^{10}\times3600=103.023\times10^{13}$
Activity after 10 hours $=\text{A}_0\text{e}^{-\lambda\text{t}}=1\times\text{e}^{\frac{-0.693}{10}\times9}=0.5\text{ci}$
No. of atoms left after 10th hour
$\text{A}_{10}=\lambda\text{N}_{10}$
$\Rightarrow\text{N}_{10}=\frac{\text{A}_{10}}{\lambda}=\frac{0.5\times3.7\times10^{10}\times3600}{\frac{0.693}{10}}$
$=26.37\times10^{10}\times3600=96.103\times10^{13}$
No.of disintegrations $=(103.023-96.103)\times10^{13}=6.92\times10^{13}$
View full question & answer→Question 533 Marks
Find the binding energy per nucleon of $\text{ }^{197}_{79}\text{Au}$ if its atomic mass is $196.96u.$
Answer$B = (Zm_p + Nm_n - M)C^2$
$Z = 79; N = 118; m_p = 1.007276u;$
$M = 196.96u; m_n = 1.008665u$
$B = [(79 \times 1.007276 + 118 \times 1.008665)u - Mu]c^2$
$= 198.597274 \times 931 - 196.96 \times 931 = 1524.302094$
so, Binding Energy per nucleon $=\frac{1524.3}{197}=7.737.$
View full question & answer→Question 543 Marks
Does a nucleus lose mass when it suffers gamma decay?
AnswerGamma rays consist of photons that are produced when a nucleus from its excited state comes to its ground state releasing energy. Since gamma rays are chargeless and massless particles, the nucleus does not suffer any loss in mass during the gamma decay.
View full question & answer→Question 553 Marks
Explain how radioactive nuclei can emit $\beta-$particles even though atomic nuclei do not contain these particles? Hence explain why the mass number of radioactive nuclide does not change during $\beta-$decay?
AnswerRadioactive nuclei do not contain electrons ($\beta-$particles), but $\beta-$particles are formed due to conversion of a neutron into a proton according to equation.
$^1_0\text{n}\ \rightarrow \ ^1_1\text{p}\ +\ _{-1} ^{0}\beta \ + \ \bar{ \upsilon}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{\beta-\text{particle}}\ \ ^\text{antincutrino}$
The $\beta-$particle so formed is emitted at once. In this process one neutron is converted into one proton; so that the number of nucleons in the nucleus remains unchanged; hence mass number of the nucleus does not change during a $\beta-$decay.
View full question & answer→Question 563 Marks
A radioactive isotope is being produced at a constant rate $\frac{\text{dN}}{\text{dt}}=\text{R}$ in an experiment. The isotope has a half-life $\text{t}_{\frac{1}2{}}.$ Show that after a time $\text{t}>>\text{t}_{\frac{1}{2}},$ the number of active nuclei will become constant. Find the value of this constant.
AnswerGiven: Half life period $=\text{t}_{\frac{1}{2}}$
Rate of radio active decay $=\frac{\text{dN}}{\text{dt}}=\text{R}\Rightarrow\text{R}=\frac{\text{dN}}{\text{dt}}$
Given after time $\text{t}>>\text{t}_{\frac{1}{2}},$ the number of active nuclei will become constant.
i.e. $\Big(\frac{\text{dN}}{\text{dt}}\Big)_{\text{present}}=\text{R}=\Big(\frac{\text{dN}}{\text{dt}}\Big)_{\text{decay}}$
$\therefore\text{R}=\Big(\frac{\text{dn}}{\text{dt}}\Big)_{\text{decay}}$
$\Rightarrow\text{R}=\lambda\text{N}$ [where, $\lambda$ = Radioactive decay constant, N = constant number]
$\Rightarrow\text{R}=\frac{0.693}{\text{t}_{\frac{1}{2}}}(\text{N})\Rightarrow\text{Rt}_{\frac{1}{2}}=0.693\text{N}\Rightarrow\text{N}=\frac{\text{Rt}_{\frac{1}{2}}}{0.693}$
View full question & answer→Question 573 Marks
Find the energy liberated in the reaction :
$^{223}Ra \rightarrow ^{209}Pb + ^{14}C.$
The atomic masses needed are as follows.
| $^{223}Ra$ |
$^{209}Pb$ |
$^{14}C$ |
| $22.018u$ |
$208.981u$ |
$14.003u$ |
Answer$^{223}Ra = 223.018u; ^{209}Pb = 208.981u; ^{14}C = 14.003u.$
$^{223}Ra \rightarrow ^{209}Pb + ^{14}C$
$\Delta\text{m} =$ mass $^{223}Ra -$ mass $(^{209}Pb + ^{14}C)$
$= 223.018 - (208.981 + 14.003) = 0.034.$
Energy $= \Delta\text{M}\times\text{u} = 0.034 \times 931 = 31.65\ Me$.
View full question & answer→Question 583 Marks
Suppose we have $12$ protons and $12$ neutrons. We can assemble them to form either a $: uMg$ nucleus or two $12C$ nuclei. In which of the two cases more energy will be liberated?
AnswerIf we assemble $6$ protons and $6$ neutrons to form $^{12}C$ nucleus, $92.15 MeV ($product of mass number and binding energy per nucleon of carbon$-12) $of energy is released.
Therefore, the energy released in the formation of two carbon nuclei is $184.3\ MeV$.
On the other hand, when $12$ protons and $12$ neutrons are combined to form $a ^{24}Mg$ atom, $198.25 MeV$ of energy $($binding energy$)$ is released.
Hence, in case of $^{24}Mg$ nucleus, more energy is liberated.
View full question & answer→Question 593 Marks
Radioactive isotopes are produced in a nuclear physics experiment at a constant rate $\frac{\text{dN}}{\text{dt}}=\text{R}.$ An inductor of inductance 100mH, a resistor of resistance $100\Omega$ and a battery are connected to form a series circuit. The circuit is switched on at the instant the production of radioactive isotope starts. It is found that $\frac{\text{i}}{\text{N}}$ remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t. Find the half-life of the isotope.
Answer$\text{R}=100\Omega;\text{ L}=100\text{mH}$
After time t, $\text{i = i}_0\Big(1-\text{e}^{\frac{-\text{t}}{\text{Lr}}}\Big)\text{ N = N}_0\big(\text{e}^{-\lambda\text{t}}\big)$
$\frac{\text{i}}{\text{N}}=\frac{\text{i}_0\big(1-\text{e}^{-\frac{\text{tR}}{\text{L}}}\big)}{\text{N}_0\text{e}^{-\lambda\text{t}}}\frac{\text{i}}{\text{N}}$ is constant i.e. independent of time.
Coefficients of t are equal $-\frac{\text{R}}{\text{L}}=-\lambda\Rightarrow\frac{\text{R}}{\text{L}}=\frac{0.693}{\text{t}_{\frac{1}{2}}}$
$=\text{t}_{\frac{1}{2}}=0.693\times10^{-3}=6.93\times10^{-4}\text{sec}.$
View full question & answer→Question 603 Marks
A human body excretes $($removes by waste discharge, sweating, etc.$)$ certain materials by a law similar to radioactivity. If technetium is injected in some form in a human body, the body excretes half the amount in $24$ hours. A patient is given an injection containing $^{99}Tc$. This isotope is radioactive with a half$-$life of $6$ hours. The activity from the body just after the injection is $6\mu\text{Ci}.$ How much time will elapse before the activity falls to $3\mu\text{Ci}?$
Answer$\text{t}_{\frac{1}{2}}=24\text{h}$
$\therefore\text{t}_{\frac{1}{2}}=\frac{\text{t}_1\text{t}_2}{\text{t}_1+\text{t}_2}=\frac{24\times6}{24+6}=4.8\text{h}.$
$\text{A}_0=6\text{rci};\text{A}=3\text{rci}$
$\therefore\text{A}=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}$
$\Rightarrow3\text{rci}=\frac{6\text{rci}}{2^{\frac{\text{t}}{4.8\text{h}}}}$
$\Rightarrow\frac{\text{t}}{24.8\text{h}}=2$
$\Rightarrow\text{t}=4.8\text{h}$
View full question & answer→Question 613 Marks
A radioactive material is reduced to $\frac{1}{16}$ of its original amount in $4$ days. How much material should one begin with so that $4 \times 10^{-3}\ kg$ of the material is left after $6$ days?
Answer$\frac{\text{N}}{\text{N}_0}=\Big(\frac{1}{2}\Big)^\text{n}$
Where $\text{n}=\frac{\text{t}}{\text{T}}$ is number of halp lives.
Given $\frac{\text{N}}{\text{N}_0}=\frac{1}{16}=\Big(\frac{1}{2}\Big)^4$
$\therefore \Big(\frac{1}{2}\Big)^4=\Big(\frac{1}{2}\Big)^\text{n}$
$\text{n}=4$
$\therefore$ Given $t = 4$ days
$\therefore \frac{\text{t}}{\text{T}}=4$
$\Rightarrow$ Half life$, \text{T}=\frac{\text{t}}{4}=\frac{4}{4}=1$ day
If $m0$ is initial mass of radioactive material, then $=\frac{\text{m}}{\text{m}_0}=\Big(\frac{1}{2}\Big)^\text{n}.$
Here $\text{n}=\frac{\text{t}}{\text{T}}=\frac{6}{1}=6,\ \text{m}=4\times 10^{-3}\text{kg}$
$\therefore \frac{\text{m}}{\text{m}_0}=\Big(\frac{1}{2}\Big)^6=\frac{1}{64}$
$m_0 = 64m = 64 \times 4 \times 10^{-3}kg$
$= 0.256\ kg$
View full question & answer→Question 623 Marks
If both the number of protons and number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction?
AnswerIn fact the number of protons and number of neutrons are same before and after a nuclear reaction, but the binding energies of nuclei present before and after a nuclear reaction are different. This difference is called the mass defect. This mass defect appears as energy of reaction. In this sense a nuclear reaction is an example of mass-energy interconversion.
View full question & answer→Question 633 Marks
Explain with example, whether the neutron-proton ratio in a nucleus increases or decreases due to $\beta-$decay.
AnswerIn $\beta-$decay a neutron is converted into a proton, so the neutron-proton ratio decreases. Equation of $\beta-$decay is:
$\text{zX}^\text{A}\rightarrow\text{Z}+_1\text{Y}^\text{A}+_{-1}\beta^0+\bar{\text{v}}$
$_{90}\text{Th}^{234}\ \rightarrow\ _{91}\text{Pa}^{234}+_{-1}\beta^0+\bar{\text{v}}$
Neutron to proton ratio before $\beta-$decay $=\frac{234-91}{90}=\frac{144}{90}=1.60$
Neutron to proton ratio aftere $\beta-$decay $=\frac{234-91}{91}=\frac{143}{91}=1.57$
$\frac{143}{91}<\frac{144}{90},$ so neutron to proton ratio in $\beta-$decay decreases.
View full question & answer→Question 643 Marks
Draw a graph showing the variation of decay rate with number of active nuclei.
AnswerAccording to Rutherford and Soddy law for redioactive decay $=\frac{-\text{dN}}{\text{dt}}=\lambda\text{N}$ where decay constant $(\lambda)$ is constant for a given radioactive material. Therefore, graph between N and $\frac{\text{dN}}{\text{dt}}$ is a straight line as shown in the diagram.
View full question & answer→Question 653 Marks
Define half-life of a radioactive sample. Which of the following radiations: $\alpha-$rays, $\beta-$rays and $\gamma-$rays.
- Are similar to X-rays.
- Are easily absorbed by matter.
- Travel with the greatest speed.
- Are similar in nature to cathode rays?
AnswerHalf-life: The half-life of a radioactive sample is defined as the time in which the mass of sample is left one half of the original mass.
- $\gamma-$rays are similar to X-rays.
- $\alpha-$rays are easily absorbed by matter.
- $\gamma-$rays travel with greatest speed.
- $\beta-$rays are similar to cathode rays.
View full question & answer→Question 663 Marks
Assume that the mass of a nucleus is approximately given by $M = Am_p$ where $A$ is the mass number. Estimate the density of matter in $\ kg/m^3$ inside a nucleus. What is the specific gravity of nuclear matter?
Answer$\text{M = Am}_{\text{p}},\text{f}=\frac{\text{M}}{\text{V}},\text{m}_{\text{p}}=1.007276\text{u}$
$\text{R = R}_0\text{A}^{\frac{1}{3}}=1.1\times10^{-15}\text{A}^{\frac{1}{3}},\\\text{u}=1.6605402\times10^{-27}\text{kg}$
$=\frac{\text{A}\times1.007276\times1.6605402\times10^{-27}}{\frac{4}{3}\times3.14\times\text{R}^3}$
$=0.300159\times10^{18}=3\times10^{17}\text{kg/m}^3.$
$‘f\ ’$ in $\text{CGS} =$ Specific gravity $=3\times10^{14}.$
View full question & answer→Question 673 Marks
How much energy is released in the following reaction?
$\text{ }^7\text{Li + p}\rightarrow\alpha+\alpha.$
Atomic mass of $^7Li = 7.0160u$ and that of $^4He = 4.0026u.$
Answer$\text{Li}^7+\text{p}\rightarrow\text{l}+\alpha+\text{E};\text{Li}^7=7.016\text{u}$
$\alpha=\text{ }^4\text{He}=4.0026\text{u};\text{p}=1.007276\text{u}$
$\text{E}=\text{Li}^7+\text{P}-2\alpha=(7.016+1.007276)\text{u}$
$-(2\times4.0026)\text{u}=0.018076\text{u}.$
$\Rightarrow0.018076\times931=16.828=16.83\ \text{MeV}.$
View full question & answer→Question 683 Marks
Calculate the $Q-$value of the fusion reaction
$^4He + ^4He = ^8Be.$
Is such a fusion energetically favourable? Atomic mass of $^8Be$ is $8.0053u$ and that of $^4He$ is $4.0026u$.
Answer$^4H + ^4H \rightarrow ^8Be$
$M(^2H) \rightarrow 4.0026u$
$M(^8Be) \rightarrow 8.0053u$
$Q$ value $= [2M(^2H) - M(^8Be)] = (2 \times 4.0026 - 8.0053)u$
$= -0.0001u = -0.0931\ Mev = -93.1\ Kev.$
View full question & answer→Question 693 Marks
Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme :
$\text{ }^{12}\text{N}\rightarrow{ }^{12}\text{C}^*+\text{e}^++\text{v}$
$\text{ }^{12}\text{C}^*\rightarrow\text{ }^{12}\text{C}+\gamma(4.43\ \text{MeV}).$
The atomic mass of $^{12}N$ is $12.018613u$.
AnswerGiven:
Atomic mass of $^{12}N, m(^{12}N) = 12.018613u$
$^{12}N \rightarrow ^{12}C* + e^+ + v$
$^{12}C* \rightarrow ^{12}C + Y (4.43\ MeV)$
Net reaction is given by
$^{12}N \rightarrow ^{12}C + e^+ + v + Y (4.43MeV)$
$Q_{value}$ of the $\beta^+$ decay will be
$Q_{value}= [m(^{12}N) - (m(^{12}C*) + 2m_e)]c^2$
$= [12.018613 \times 931\ MeV - (12 \times 931 + 4.43)\ MeV - (2 \times 511)\ keV]$
$= [11189.3287 - 11176.43 - 1.022]\ MeV$
$= 11.8767MeV = 11.88\ MeV$
The maximum kinetic energy of beta particle will be $11.88\ MeV,$ assuming that neutrinos have zero energy.
View full question & answer→Question 703 Marks
In a typical fission reaction, the nucleus is split into two middle-weight nuclei of unequal masses. Which of the two (heavier or lighter) has greater kinetic energy? Greater linear momentum?
AnswerTwo photons having equal liner momentum have equal wavelengths as here for both the photons the direction and magnitude of linear momentum will be same. For the rest of the options, magnitude will be same but nothing can be said about the direction of the photons.
Hence the correct option is D.
View full question & answer→Question 713 Marks
A molecule. of hydrogen contains two protons and two electrons. The nuclear force between these two protons is always neglected while discussing the behaviour of a hydrogen molecule. Why?
AnswerInside the nucleus, two protons exert nuclear force on each other. These forces are short-ranged (a few fm), strong and attractive forces. They also exert electrostatic repulsive force (long-ranged). While discussing the behaviour of a hydrogen molecule, the nuclear force between the two protons is always neglected. This is because the separation between the two protons in the molecule is ~70pm which is much greater than the range of the nuclear force.
View full question & answer→Question 723 Marks
A certain sample of a radioactive material decays at the rate of 500 per second at a certain time. The count rate falls to 200 per second after 50 minutes.
- What is the decay constant of the sample?
- What is its half-life?
Answer$\text{A = 200, A}_0 = 500, \text{t = 50 min}$
- $\text{A = A}_0\text{e}^{-\lambda\text{t}}$
$200=500\times\text{e}^{-50\times60\times\lambda}$
$\Rightarrow\lambda=3.05\times10^{-4}\text{s}.$
- $\text{t}_{\frac{1}{2}}=\frac{0693}{\lambda}=\frac{0.693}{0.000305}=2272.13\sec=38\text{min}$
View full question & answer→Question 733 Marks
Prove that the instantaneous rate of change of the activity of a radioactive substance is inversely proportional to the square of its half-life.
AnswerActivity of a radioactive substance.
$\text{R}\Big(=-\frac{\text{dN}}{\text{dt}}\Big)=\lambda\text{N}\ \dots(\text{i})$
Rate of change of activity
$\frac{\text{dR}}{\text{dt}}=\lambda \Big(\frac{\text{dN}}{\text{dt}}\Big)=\lambda(-\lambda\text{N})$
$=-\lambda^2\text{N}$
As $\lambda =\frac{\log_\text{e}2}{\text{T}_{\frac{1}{2}}}$ $\therefore \frac{\text{dR}}{\text{dt}}=-\Big(\frac{\log_\text{e}2}{\text{T}_{\frac{1}{2}}}\Big)^2\text{N}$
$\therefore$ Instantaneous activity, $\frac{\text{dR}}{\text{dt}}\propto\frac{1}{\text{T}^2_{\frac{1}{2}}}$
View full question & answer→Question 743 Marks
A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life $\tau.$ Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.
Answer$\text{Q = qe}^{\frac{-\text{t}}{\text{CR}}};\text{A = A}_0\text{e}^{-\lambda\text{t}}$
$\frac{\text{Energy}}{\text{Activity}}=\frac{1\text{q}^2\times\text{e}^{\frac{-2\text{t}}{\text{CR}}}}{2\text{CA}_0\text{e}^{-\lambda\text{t}}}$
Since the term is independent of time, so their coefficients can be equated,
So, $\frac{2\text{t}}{\text{CR}}=\lambda\text{t}$
$\lambda=\frac{2}{\text{CR}}$
$\frac{1}{\tau}=\frac{2}{\text{CR}}$
$\text{R}=2\frac{\tau}{\text{C}}$
View full question & answer→Question 753 Marks
With the help of an example, explain how the neutron to proton ratio changes during $\alpha-$decay of a nucleus.
AnswerLet us like the example of $\alpha-$decay of $^{238}_{92}\text{U}.$ The decay scheme is
$^{238}_{92}\text{U}\rightarrow ^{234}_{90}\text{Th}+^4_2\alpha \text{ (or}^4_2\text{He})$
Neutron to proton ratio before $\alpha-$decay $=\frac{238-92}{92}=\frac{146}{92}=1.59$
Neutron to proton ratio after $\alpha-$decay $=\frac{238-90}{90}=\frac{144}{90}=1.60$
$\frac{146}{92}<\frac{144}{90}$
This shows that the neutron to proton ratio increases during $\alpha-$decay of a nucleus.
View full question & answer→Question 763 Marks
Calculate the energy that can be obtained from $1\ kg$ of water through the fusion reaction
$^2H + ^2H \rightarrow ^3H + p.$
Assume that $1.5 \times 10^{-2}\%$ of natural water is heavy water $D_2O ($by number of molecules$)$ and all the deuterium is used for fusion.
AnswerGiven:
$18g$ of water contains $6.023 \times 10^{23 }$ molecules.
$\therefore1000\text{g}$ of water $=\frac{6.023\times10^{23}\times1000}{18}=3.346\times10^{25}$ molecules
$\%$ of deuterium $=3.346\times10^{25}\times\frac{0.015}{100}=0.05019\times10^{23}$
Energy of deuterium $=30.4486\times10^{25}$
$= [2 \times m(^2H) - m(^3H) - m_p]c^2$
$= (2 \times 2.014102u - 3.016049u - 1.007276u)c^2$
$= 0.004879 \times 931Me$
$= 4.542349Me$
$= 7.262 \times 10^{-13}J$
Total energy $= 0.05019 \times 10^{23} \times 7.262 \times 10^{-13}J$
$= 3644MJ$
View full question & answer→Question 773 Marks
$^{228}Th$ emits an alpha particle to reduce to $^{224}Ra.$ Calculate the kinetic energy of the alpha particle emitted in the following decay:
$\text{ }^{228}\text{Th}\rightarrow\text{ }^{224}\text{Ra}^*+\alpha$
$\text{ }^{224}\text{Ra}^*\rightarrow\text{ }^{224}\text{Ra}+\gamma(217\text{kev}).$
Atomic mass of $^{228}Th$ is $228.028726u,$ that of $^{224}Ra$ is $224.020196u$ and that of $\text{ }^4_2\text{He}$ is $4.00260u.$
AnswerMass $\text{ }^{228}\text{Th}=228.028726\text{u};\text{ }^{224}\text{Ra}=224.020196\text{u}$
$\alpha=\text{ }^4_2\text{He}\rightarrow4.00260\text{u}$
$\text{ }^{228}\text{Th}\rightarrow\text{ }^{224}\text{Ra}^*+\alpha$
$\text{ }^{224}\text{Ra}^*\rightarrow\text{ }^{224}\text{Ra + v}(217\text{kev})$
Now, Mass of $\text{ }^{224}\text{Ra}^* = 224.020196 \times 931 + 0.217\text{ Mev} $
$= 208563.0195\text{Mev.}$
$KE$ of $\alpha=\text{E}^{226}\text{Th}-\text{E}(\text{ }^{224}\text{Ra}^*+\alpha)$
$= 228.028726\times 931-[208563.0195 + 4.00260\times931]$
$= 5.30383\text{Mev}= 5.304\text{Mev.}$
View full question & answer→Question 783 Marks
Carbon $(Z = 6)$ with mass number $11$ decays to boron $(Z = 5)$.
- Is it a $\beta^+-$decay or a $\beta^--$decay?
- The half$-$life of the decay scheme is $20.3$ minutes. How much time will elapse before a mixture of $90\%$ carbon$-11$ and $10\%$ boron$-11\ ($by the number of atoms$)$ converts itself into a mixture of $10\%$ carbon$-11$ and $90\%$ boron$-11$?
Answer$\text{P}\rightarrow\text{n + e}^++\text{v}$
Hence it is a $\beta^+$ decay.
Let the total no. of atoms be $100 N_0$.
| |
Carbon |
Boron |
| Initially |
$90N_0$ |
$10N_0$ |
| Finally |
$10N_0$ |
$90N_0$ |
Now, $10\text{N}_0=90\text{N}_0\text{e}^{-\lambda\text{t}}$
$\Rightarrow\frac{1}9{}=\text{e}^{\frac{-0.693}{20.3}\times\text{t}}\ \Big[$because $\text{t}_{\frac{1}2{}}=20.3\ \text{min}\Big]$
$\Rightarrow$ In $\frac{1}9{}=\frac{-0.693}{20.3}\text{t}$
$\Rightarrow\text{t}=\frac{2.1972\times20.3}{0.693}=64.36=64\ \text{min}.$ View full question & answer→Question 793 Marks
Consider a radioactive nucleus $A$ which decays to a stable nucleus $C$ through the following sequence$: A \rightarrow B \rightarrow C$
Here $B$ is an intermediate nuclei which is also radioactive. Considering that there are $N_0$ atoms of $A$ initially, plot the graph showing the variation of number of atoms of $A$ and $B$ versus time.
AnswerConsider the situation shown in the graph.
At $t = 0, NA = NO$ while $NB = 0.$ As time increases, $NA$ falls off exponentially, the number of atoms of $B$ increases, becomes maximum and finally decays to zero at $\infty ($following exponential decay law$).$ View full question & answer→Question 803 Marks
A small telescope has an objective lens of focal length $144 \ cm$ and an eyepiece of focal length $6.0 \ cm$. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
AnswerFocal length of the objective lens$, f_0 = 144 \ cm$
Focal length of the eyepiece$, f_e = 6.0 \ cm$
The magnifying power of the telescope is given as:
$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}$
$=\frac{144}{6}=24$
The separation between the objective lens and the eyepiece is calculated as:
$f_0 + f_e$
$= 144 + 6 = 150 \ cm$
Hence, the magnifying power of the telescope is $24$ and the separation between the objective lens and the eyepiece is $150 \ cm$.
View full question & answer→Question 813 Marks
Draw the graph showing the variation of binding energy per nucleon with mass numbers. Give the reason for the decrease of binding energy per nucleon for nuclei with higher mass number.
AnswerThe graph of the binding energy per nucleon versus mass number A is shown in figure. The decrease of the binding energy per nucleon for nuclei with high mass number is due to increased coulomb repulsion between protons inside the nucleus.
View full question & answer→Question 823 Marks
Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5kT equals the Coulomb potential energy at 2fm.
Answer$\text{PE}=\frac{\text{Kq}_1\text{q}_2}{\text{r}}=\frac{9\times10^9\times(2\times1.6\times10^{-19})^2}{\text{r}} \ ...(1)$
$1.5\text{KT}=1.5\times1.38\times10^{-23}\times\text{T} \ ...(2)$
Equating (1) and (2) $1.5\times1.38\times10^{-23}\times\text{T}=\frac{9\times10^9\times10.24\times10^{-38}}{2\times10^{-15}}$
$\Rightarrow\text{T}=\frac{9\times10.24\times10^{-38}}{2\times10^{-15}\times1.5\times1.38\times10^{-23}}$
$=22.26087\times10^9\text{K}=2.23\times10^{10}\text{K}$
View full question & answer→Question 833 Marks
The decay constant of $^{238}U$ is $4.9 \times 10^{-18} S^{-1}.$
- What is the average$-$life of $^{238}U$?
- What is the half$-$life of $^{238}U$?
- By what factor does the activity of a $^{238}U$ sample decrease in $9 \times 10^9$ years?
Answer$\lambda=4.9\times10^{-18}\text{s}^{-1}$
- Avg. life of $\text{ }^{238}\text{U}=\frac{1}{\lambda}=\frac{1}{4.9\times10^{-18}}=\frac{1}{4.9}\times10^{-18}\sec.$
$=6.47\times10^{3}\text{years}.$
- Half life of uranium $=\frac{0.693}{\lambda}=\frac{0.693}{4.9\times10^{-18}}=4.5\times10^9$ years.
- $\text{A}=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}$
$\Rightarrow\frac{\text{A}_0}{\text{A}}=2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=2^2=4.$ View full question & answer→Question 843 Marks
When charcoal is prepared from a living tree, it shows a disintegration rate of $15.3$ disintegrations of $^{14}C$ per gram per minute. A sample from an ancient piece of charcoal shows $^{14}C$ activity to be $12.3$ disintegrations per gram per minute. How old is this sample? Half$-$life of $^{14}C$ is $5730y.$
Answer$\text{A}_0=15.3;\text{ A}=12.3;\text{t}_{\frac{1}{2}}=5730$ year
$\lambda=\frac{0.6931}{\text{T}_{\frac{1}{2}}}=\frac{0.6931}{5730}\text{yr}^{-1}$
Let the time passed be $t,$
We know $\text{A = A}_0\text{e}^{-\lambda\text{t}}-\frac{0.6931}{5730}\times\text{t}$
$\Rightarrow12.3=15.3\times\text{e}$
$\Rightarrow\text{t}=1804.3$ years.
View full question & answer→Question 853 Marks
Is it easier to take out a nucleon from carbon or from iron? Fi-om iron or from lead?
AnswerBinding energy per nucleon of a nucleus is defined as the energy required to break-off a nucleon from it.
- As the binding energy per nucleon of iron is more than that of carbon, it is easier to take out a nucleon from carbon than iron.
- As the binding energy per nucleon of iron is more than that of lead. Therefore, it is easier to take out a nucleon from lead as compared to iron.
View full question & answer→Question 863 Marks
radioactive isotope has a half-life of 5 years. After how much time is its activity reduced to 3.125% of its original activity?
AnswerWe know $\frac{\text{R}}{\text{R}}_0=\Big(\frac{1}{2}\Big)^\text{}n$
Given $\frac{\text{R}}{\text{R}}_0=3.125\%=\frac{3.125}{100}$
$\therefore \frac{3.125}{100}=\Big(\frac{1}{2}\Big)^\text{n}$
$\frac{1}{32}=\Big(\frac{1}{2}\Big)^\text{n}$
$\Big(\frac{1}{2}\Big)^5=\Big(\frac{1}{2}\Big)^\text{n}$
$\Rightarrow \text{n}=5$
Given T = 5 years
As $\text{n}=\frac{\text{t}}{\text{T}}$
$\therefore \frac{\text{t}}{\text{T}}=5$
$\text{t}=5\times5=25\text{years}$
View full question & answer→Question 873 Marks
Distinguish between isotopes and isobars. Give one example for each of the species.
Answer
| S. No. |
Isotopes |
Isobars |
| 1. |
The nuclides having the same atomic number (Z). |
The nuclides having the same atomic mass (A) but. |
| 2. |
But different atomic masses (A) are called isotopes. Examples: $^1_1\text{H},\ ^2_1\text{H},\ ^3_1\text{H}$ |
Different atomic numbers (Z) are called isobars. Examples: $^3_1\text{H},\ ^3_2\text{He}$ |
View full question & answer→Question 883 Marks
$\text{ }^{197}_{80}\text{Hg}$ decay to $\text{ }^{197}_{79}\text{Au}$ through electron capture with a decay constant of 0.257 per day.
- What other particle or particles are emitted in the decay?
- Assume that the electron is captured from the K shell. Use Moseley's law $\sqrt{\text{v}}=\text{a(Z}-\text{b})$ with $\text{a}=4.95\times10^7\text{s}^{-\frac{1}{2}}$ and b = 1 to find the wavelength of the $\text{K}_{\alpha}$ X-ray emitted following the electron capture.
Answer
- $\text{P + e}\rightarrow\text{n + v}$ neutrino $\big[\text{a}\rightarrow4.95\times10^7\text{s}^{-\frac{1}{2}};\text{b}\rightarrow1\big]$
- $\sqrt{\text{f}}=\text{a(z}-\text{b})$
$\Rightarrow\sqrt{\frac{\text{c}}{\lambda}}=4.95\times10^7(79-1)=4.95\times10^7\times78$
$\Rightarrow\frac{\text{c}}{\lambda}=(4.95\times78)^2\times10^{14}$
$\Rightarrow\lambda=\frac{3\times10^8}{14903.2\times10^{14}}$
$=2\times10^{-5}\times10^{-6}=2\times10^{-4}\text{m}=20\text{pm}$ View full question & answer→Question 893 Marks
The count rate from a radioactive sample falls from $4.0 \times 10^6$ per second to $1.0 \times 10^{6 }$ per second in $20$ hours. What will be the count rate $100$ hours after the beginning?
Answer$\text{A}_0=4\times10^5$ disintegration/sec $\text{A}'=1\times10^6\ dis/\sec;\ t = 20$ hours.
$\text{A}'=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}$
$\Rightarrow2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=\frac{\text{A}_0}{\text{A}'}$
$\Rightarrow2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=4$
$\Rightarrow\frac{\text{t}}{\text{t}_{\frac{1}{2}}}=2$
$\Rightarrow\text{t}^{\frac{1}{2}}=\frac{\text{t}}{2}=\frac{20\text{ hours}}{2}=10\text{ hours}.$
$\text{A}''=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}$
$\Rightarrow\text{A}''=\frac{4\times10^6}{2^{\frac{100}{10}}}$
$=0.00390625\times10^6$
$=3.9\times10^3$ dintegrations$/\sec.$
View full question & answer→Question 903 Marks
Which sample, A or B shown in Fig. has shorter mean-life?
AnswerB has shorter mean life as $\lambda$ is greater for B. This can be explained mathematically as given below
From the given graph, at $\text{t}=0,\Big(\frac{\text{dN}}{\text{dt}}\Big)_\text{A}=\Big(\frac{\text{dN}}{\text{dt}}\Big)_\text{B}\Rightarrow\ (\text{N}_0)_\text{A}=(\text{N}_0)_\text{B}$
Considering any instant t by drawing a line perpendicular to time axis, we find that $\Big(\frac{\text{dN}}{\text{dt}}\Big)_\text{A}>\Big(\frac{\text{dN}}{\text{dt}}\Big)_\text{B}$
$\Rightarrow\ \lambda_\text{A}\text{N}_\text{A}>\lambda_\text{B}\text{N}_\text{B}$
$\because\ \text{N}_\text{A}>\text{N}_\text{B}$ (rate of decay of B is slower)
$\because\ \lambda_\text{B}>\lambda_\text{A}$
As, average life, $\tau=\frac{1}{\lambda}$
$\Rightarrow\ \tau_\text{A}>\tau_\text{B}$
View full question & answer→Question 913 Marks
The drift current in a reverse-biased p-n junction increases in magnitude if the temperatu,re of the junction is increased. Explain this on the basis of creation of hole-electron pairs.
AnswerWhen the temperature of a reverse-biassed p‒n junction is increased, the breaking of bonds takes place because of the increase in the thermal energy of the charge carriers. Drift current is due to the flow of the minority carriers across the junction. So, when a p‒n junction is reverse biassed, the applied voltage supports the flow of minority charge carriers across the junction. Thus, the drift current increases with increase in temperature in a reverse biassed p‒n junction.
View full question & answer→Question 923 Marks
Consider the situation of the previous problem. Suppose the production of the radioactive isotope starts at $t = 0$. Find the number of active nuclei at time $t$.
AnswerLet $N_0 =$ No. of radioactive particle present at time $t = 0$
$N =$ No. of radio active particle present at time $t$.
$\therefore\text{N = N}_0\text{e}^{-\lambda\text{t}}$ [$\lambda$ -Radioactive decay constant$]$
$\therefore$ The no.of particles decay $=\text{N}_0-\text{N}=\text{N}_0-\text{N}_0\text{e}^{-\lambda\text{t}}=\text{N}_0(1-\text{e}^{-\lambda\text{t}})$
We know, $\text{A}_0=\lambda\text{N}_0;\text{R}=\lambda\text{N}_0;\text{N}_0=\frac{\text{R}}{\lambda}$
From the above equation
$\text{N = N}_0(1-\text{e}^{-\lambda\text{t}})=\frac{\text{R}}{\lambda}(1-\text{e}^{-\lambda\text{t}})\ ($substituting the value of $N_0)$
View full question & answer→Question 933 Marks
Consider two pairs of neutrons. In each pair, the separation between the neutrons is the same. Can the force between the neutrons have different magnitudes for the two pairs?
AnswerNeutrons are chargeless particles and they exert only short range nuclear forces on each other. If we have two pairs of neutrons and the separation between them is same in both the pairs. The force between the neutrons will be of same magnitude for the two pairs until there is some other influence on any of them.
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