Question 13 Marks
Calculate the energy released by 1g of natural uranium assuming 200MeV is released in each fission event and that the fissionable isotope $^{235}U$ has an abundance of 0.7% by weight in natural uranium.
Answer1g of ‘I’ contain 0.007g $U^{235}$
So, 235g contains 6.023 × 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 23 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 33 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 43 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 53 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 63 Marks
Find the binding energy per nucleon of $\text{ }^{197}_{79}\text{Au}$ if its atomic mass is 196.96u.
Answer$B=\left(Z m_p+N m_n-M\right) C^2$
$Z=79 ; N=118 ; m_p=1.007276 u ; M=196.96 u ; m_n=1.008665 u$
$B=[(79 \times 1.007276+118 \times 1.008665) u-M u] c^2$
$=198.597274 \times 931-196.96 \times 931=1524.302094$
so, Binding Energy per nucleon $=\frac{1524.3}{197}=7.737$.
View full question & answer→Question 73 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 83 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 93 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 103 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} \mathrm{Ra}=223.018 \mathrm{u} ;{ }^{209} \mathrm{~Pb}=208.981 \mathrm{u} ;{ }^{14} \mathrm{C}=14.003 \mathrm{u} .{ }^{223} \mathrm{Ra} \rightarrow{ }^{209} \mathrm{~Pb}$
$+{ }^{14} \mathrm{C} \Delta \mathrm{m}=\text { mass }{ }^{223} \mathrm{Ra}-\operatorname{mass}\left({ }^{209} \mathrm{~Pb}+{ }^{14} \mathrm{C}\right)$
$ =223.018-(208.981+14.003)=0.034 . \text { Energy }=\Delta \mathrm{M} \times \mathrm{u}=0.034 \times 931=31.65 \mathrm{Me} .$
View full question & answer→Question 113 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 123 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 133 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 143 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 153 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 163 Marks
Assume that the mass of a nucleus is approximately given by $M=A m_p$ where $A$ is the mass number. Estimate the density of matter in $\mathrm{kg} / \mathrm{m}^3$ 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}$$\text{R = R}_0\text{A}^{\frac{1}{3}}=1.1\times10^{-15}\text{A}^{\frac{1}{3}},\\\text{u}=1.6605402\times10^{-27}\text{kg}$
$=\frac{\text{A}\times1.007276\times1.6605402\times10^{-27}}{\frac{4}{3}\times3.14\times\text{R}^3}$
$=0.300159\times10^{18}=3\times10^{17}\text{kg/m}^3.$
‘f’ in CGS = Specific gravity $=3\times10^{14}.$
View full question & answer→Question 173 Marks
$\text{ }^{197}_{80}\text{Hg}$ decay to $\text{ }^{197}_{79}\text{Au}$ through electron capture with a decay constant of $0.257$ per day.
- What other particle or particles are emitted in the decay?
- Assume that the electron is captured from the $K$ shell. Use Moseley's law $\sqrt{\text{v}}=\text{a(Z}-\text{b})$ with $\text{a}=4.95\times10^7\text{s}^{-\frac{1}{2}}$ and $b = 1$ to find the wavelength of the $\text{K}_{\alpha} X-$ray emitted following the electron capture.
Answer
- $\text{P + e}\rightarrow\text{n + v}$ neutrino $\big[\text{a}\rightarrow4.95\times10^7\text{s}^{-\frac{1}{2}};\text{b}\rightarrow1\big]$
- $\sqrt{\text{f}}=\text{a(z}-\text{b})$
$\Rightarrow\sqrt{\frac{\text{c}}{\lambda}}=4.95\times10^7(79-1)=4.95\times10^7\times78$
$\Rightarrow\frac{\text{c}}{\lambda}=(4.95\times78)^2\times10^{14}$
$\Rightarrow\lambda=\frac{3\times10^8}{14903.2\times10^{14}}$
$=2\times10^{-5}\times10^{-6}$
$=2\times10^{-4}\text{m}$
$=20\text{pm}$ View full question & answer→Question 183 Marks
How much energy is released in the following reaction?$\text{ }^7\text{Li + p}\rightarrow\alpha+\alpha.$
Atomic mass of ${ }^7 \mathrm{Li}=7.0160 \mathrm{u}$ and that of ${ }^4 \mathrm{He}=4.0026 \mathrm{u}$.
Answer$\text{Li}^7+\text{p}\rightarrow\text{l}+\alpha+\text{E};\text{Li}^7=7.016\text{u}$$\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 193 Marks
Calculate the Q-value of the fusion reaction
${ }^4 \mathrm{He}+{ }^4 \mathrm{He}={ }^8 \mathrm{Be} .$
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 203 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 ${ }^{12} \mathrm{~N}, \mathrm{~m}\left({ }^{12} \mathrm{~N}\right)=12.018613 \mathrm{u}$
${ }^{12} \mathrm{~N} \rightarrow{ }^{12} \mathrm{C}^*+\mathrm{e}^{+}+\mathrm{v}$
${ }^{12} \mathrm{C}^{\star} \rightarrow{ }^{12} \mathrm{C}+\mathrm{Y}(4.43 \mathrm{MeV})$
Net reaction is given by
${ }^{12} \mathrm{~N} \rightarrow{ }^{12} \mathrm{C}+\mathrm{e}^{+}+\mathrm{v}+\mathrm{Y}(4.43 \mathrm{MeV})$
$\mathrm{Q}_{\text {value }}$ of the $\beta^{+}$decay will be
$\text { Qvalue }=\left[\mathrm{m}\left({ }^{12} \mathrm{~N}\right)-\left(\mathrm{m}\left({ }^{12} \mathrm{C}^{\star}\right)+2 \mathrm{me}_{\mathrm{e}}\right)\right] \mathrm{c}^2$
$=[12.018613 \times 931 \mathrm{MeV}-(12 \times 931+4.43) \mathrm{MeV}-(2 \times 511) \mathrm{keV}]$
$=[11189.3287-11176.43-1.022] \mathrm{MeV}$
$=11.8767 \mathrm{MeV}=11.88 \mathrm{MeV}$
The maximum kinetic energy of beta particle will be 11.88 MeV , assuming that neutrinos have zero energy.
View full question & answer→Question 213 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 223 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 233 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 243 Marks
Calculate the energy that can be obtained from $1 kg$ of water through the fusion reaction
${ }^2 \mathrm{H}+{ }^2 \mathrm{H} \rightarrow{ }^3 \mathrm{H}+\text { p. }$
Assume that $1.5 \times 10^{-2} \%$ of natural water is heavy water $\mathrm{D}_2 \mathrm{O}$ (by number of molecules) and all the deuterium is used for fusion.
AnswerGiven: 18 g of water contains $6.023 \times 10^{23}$ molecules. $\therefore 1000 \mathrm{~g}$ of water $=\frac{6.023 \times 10^{23} \times 1000}{18}=3.346 \times 10^{25}$ molecules % of deuterium $=3.346 \times 10^{25} \times \frac{0.015}{100}=0.05019 \times 10^{23}$ Energy of deuterium $=30.4486 \times 10^{25}=[2 \times \mathrm{m}(2 \mathrm{H})$
$-\mathrm{m}\left({ }^3 \mathrm{H}\right)-\mathrm{m}_{\mathrm{p}} \mathrm{~J} \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}=3644 \mathrm{MJ}$
View full question & answer→Question 253 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 263 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 273 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 283 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 293 Marks
The decay constant of ${ }^{238} \mathrm{U}$ is $4.9 \times 10^{-18} \mathrm{~S}^{-1}$.
a. What is the average$-$life of ${ }^{238} \mathrm{U}$ ?
b. What is the half$-$life of ${ }^{238} \mathrm{U}$ ?
c. By what factor does the activity of a ${ }^{238} \mathrm{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 303 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 $\text{A = A}_0\text{e}^{-\lambda\text{t}}-\frac{0.6931}{5730}\times\text{t}$
$\Rightarrow12.3=15.3\times\text{e}$
$\Rightarrow\text{t}=1804.3\text{ years.}$
View full question & answer→