Question 513 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 523 Marks
A uranium reactor develops thermal energy at a rate of 300MW. Calculate the amount of $^{235}U$ being consumed every second. Average released per fission is 200MeV.
AnswerLet n atoms disintegrate per second
Total energy emitted/sec $= (n \times 200 \times 106 \times 1.6 \times 10^{-19})J =$ Power
$300MW = 300 × 106$ 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 533 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 543 Marks
Find the binding energy per nucleon of $\text{ }^{197}_{79}\text{Au}$ if its atomic mass is 196.96u.
Answer$ B=\left(Z m_p+N m_n-M\right) C^2 $
$ Z=79 ; N=118 ; m_p=1.007276 u ; M=196.96 u ; m_n=1.008665 u $
$ B=[(79 \times 1.007276+118 \times 1.008665) u-M u] c^2 $
$ =198.597274 \times 931-196.96 \times 931=1524.302094 $
$ \text { so, Binding Energy per nucleon }=\frac{1524.3}{197}=7.737 .$
View full question & answer→Question 553 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 563 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 573 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 583 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 593 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 603 Marks
Find the energy liberated in the reaction:
$^{223}Ra → ^{209}Pb + ^{14}C.$
The atomic masses needed are as follows.
| $^{223}Ra$ |
$^{209}Pb$ |
$^{14}C$ |
| $22.018u$ |
$208.981u$ |
$14.003u$ |
Answer$^{223}Ra = 223.018u; ^{209}Pb = 208.981u; ^{14}C = 14.003u.$
$^{223}Ra → ^{209}Pb + ^{14}C$
$\Delta\text{m} = mass ^{223}Ra - mass(^{209}Pb + ^{14}C)$
$= 223.018 - (208.981 + 14.003) = 0.034.$
Energy = $\Delta\text{M}\times\text{u} = 0.034 × 931 = 31.65$Me.
View full question & answer→Question 613 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 623 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 ${ }^{12} \mathrm{C}$ 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 ${ }^{24} \mathrm{Mg}$ atom, 198.25 MeV of energy (binding energy) is released. Hence, in case of ${ }^{24} \mathrm{Mg}$ nucleus, more energy is liberated.
View full question & answer→Question 633 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 $4 × 10^{-3}kg$ 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
$\therefore \frac{\text{t}}{\text{T}}=4\Rightarrow \text{Half life,}\ \text{T}=\frac{\text{t}}{4}=\frac{4}{4}=1\text{day}$
If m0 is initial mass of radioactive material, then $=\frac{\text{m}}{\text{m}_0}=\Big(\frac{1}{2}\Big)^\text{n}.$
Here $\text{n}=\frac{\text{t}}{\text{T}}=\frac{6}{1}=6,\ \text{m}=4\times 10^{-3}\text{kg}$
$\therefore \frac{\text{m}}{\text{m}_0}=\Big(\frac{1}{2}\Big)^6=\frac{1}{64}$
$m_0 = 64m = 64 \times 4 \times 10^{-3}kg$
$= 0.256kg$
View full question & answer→Question 643 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 653 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 663 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 673 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.
View full question & answer→Question 683 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 693 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 703 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 713 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 $3.48 \times 10^6m$, and the radius of lunar orbit is $3.8 \times 10^8m.$
AnswerFocal length of the objective lens, $f_0 = 15 m = 15 \times 10^2cm$
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, $d = 3.48 \times 10^6m$
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 723 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 CGS = Specific gravity $=3\times10^{14}.$
View full question & answer→Question 733 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 743 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 753 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$ { }^4 \mathrm{H}+{ }^4 \mathrm{H} \rightarrow{ }^8 \mathrm{Be} $
$ \mathrm{M}\left({ }^2 \mathrm{H}\right) \rightarrow 4.0026 \mathrm{u} $
$ \mathrm{M}\left({ }^8 \mathrm{Be}\right) \rightarrow 8.0053 \mathrm{u} $
$ \mathrm{Q} \text { value }=\left[2 \mathrm{M}\left({ }^2 \mathrm{H}\right)-\mathrm{M}\left({ }^8 \mathrm{Be}\right)\right]=(2 \times 4.0026-8.0053) \mathrm{u} $
$ =-0.0001 \mathrm{u}=-0.0931 \mathrm{Mev}=-93.1 \mathrm{Kev} .$
View full question & answer→Question 763 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 $^{12}N, m(^{12}N) = 12.018613u$
$^{12}N → ^{12}C^* + e^+ + v$
$^{12}C^* → ^{12}C + Y (4.43MeV)$
Net reaction is given by
$^{12}N → ^{12}C + e^+ + v + Y (4.43MeV)$
$Q_{value}$ of the $\beta^+$ decay will be
$Q_{value}= [m(^{12}N) - (m(^{12}C^*) + 2m_e)]c^2$
$= [12.018613 × 931MeV - (12 × 931 + 4.43) MeV - (2 × 511)keV]$
$= [11189.3287 - 11176.43 - 1.022]MeV$
$= 11.8767MeV = 11.88MeV$
The maximum kinetic energy of beta particle will be 11.88MeV, assuming that neutrinos have zero energy.
View full question & answer→Question 773 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 783 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 793 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 803 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 813 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 823 Marks
Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence:
A → B → 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 ∞ (following exponential decay law).
View full question & answer→Question 833 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 843 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 853 Marks
Calculate the energy that can be obtained from 1kg of water through the fusion reaction
$^2H + ^2H → ^3H + p.$
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 $=30.4486\times10^{25}$
$ =\left[2 \times \mathrm{m}\left({ }^2 \mathrm{H}\right)-\mathrm{m}\left({ }^3 \mathrm{H}\right)-\mathrm{m}_{\mathrm{p}}\right] \mathrm{c}^2 $
$ =(2 \times 2.014102 \mathrm{u}-3.016049 \mathrm{u}-1.007276 \mathrm{u}) \mathrm{c}^2 $
$ =0.004879 \times 931 \mathrm{Me} $
$ =4.542349 \mathrm{Me} $
$ =7.262 \times 10^{-13} \mathrm{~J} $
$ \text { Total energy }=0.05019 \times 10^{23} \times 7.262 \times 10^{-13} \mathrm{~J} $
= 3644MJ
View full question & answer→Question 863 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 873 Marks
Carbon (Z = 6) with mass number 11 decays to boron (Z = 5).
- Is it a $\beta^+-\text{decay}$ or a $\beta^--\text{decay}?$
- 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 $\beta^+\text{decay}.$
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 $\text{t}_{\frac{1}2{}}=20.3\text{min}\Big]$
$\Rightarrow\text{In}\frac{1}9{}=\frac{-0.693}{20.3}\text{t}\Rightarrow\text{t}=\frac{2.1972\times20.3}{0.693}=64.36=64\text{min}.$ View full question & answer→Question 883 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 893 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 903 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 913 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 923 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 $=\frac{0.693}{\lambda}=\frac{0.693}{4.9\times10^{-18}}=4.5\times10^9\text{years}.$
- $\text{A}=\frac{\text{A}_0}{2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}}\Rightarrow\frac{\text{A}_0}{\text{A}}=2^{\frac{\text{t}}{\text{t}_{\frac{1}{2}}}}=2^2=4.$
View full question & answer→Question 933 Marks
When charcoal is prepared from a living tree, it shows a disintegration rate of 15.3 disintegrations of ${ }^{14} \mathrm{C}$ per gram per minute. A sample from an ancient piece of charcoal shows ${ }^{14} \mathrm{C}$ activity to be 12.3 disintegrations per gram per minute. How old is this sample? Half-life of ${ }^{14} \mathrm{C}$ is 5730 y .
Answer$\text{A}_0=15.3;\text{ A}=12.3;\text{t}_{\frac{1}{2}}=5730\text{ year}$
$\lambda=\frac{0.6931}{\text{T}_{\frac{1}{2}}}=\frac{0.6931}{5730}\text{yr}^{-1}$
Let the time passed be t,
We know $\text{A = A}_0\text{e}^{-\lambda\text{t}}-\frac{0.6931}{5730}\times\text{t}$
$\Rightarrow12.3=15.3\times\text{e}$
$\Rightarrow\text{t}=1804.3\text{ years.}$
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