Concept Of Physics The Nucleus

H C Verma

Concept Of Physics

1.   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 ?

$M = Amp,\ f = \dfrac{M}{V},\ mp = 1.007276\ u$ $\\$ $R = R_0A^\dfrac{1}{3} = 1.1 \times 10^{–15}\ A^\dfrac{1}{3},\ u = 1.6605402 \times 10^{–27} kg$ $\\$ $= \dfrac{A \times 1.007276 \times 1.6605402 \times 10^{-27}}{\dfrac{4}{3} \times 3.14 \times R^3} = 0.300159 \times 10^{18} = 3 \times 10^{17}\ kg/m^3.$ $\\$ ‘$f’$ in $CGS$ = Specific gravity $= 3 \times 10^{14}$.

2.   A neutron, star has a density equal to that of the nuclear matter. Assuming the star to be spherical, find the radius of a neutron star whose mass is $4\cdot0 \times 10\ kg$ (twice the mass of the sun).

$f = {M}{v} \Rightarrow V = \dfrac{M}{f} = \dfrac{4 \times 10^{30}}{2.4 \times 10^{17}}= \dfrac{1}{0.6} \times 10^{13} \dfrac{1}{6} \times 10^{14}$ $\\$ $V = \dfrac{4}{3} \pi \times R^3$. $\\$ $\Rightarrow \dfrac{1}{6}\times 10^{14} = \dfrac{4}{3} \pi \times R^3 \times R^3 = {1}{6}\times \dfrac{3}{4} \times \dfrac{1}{\pi} \times 10^{14}$ $\\$ $\Rightarrow R^3 = \dfrac{1}{8} \times \dfrac{100}{\pi} \times 10^{12}$ $\\$ $\therefore R = \dfrac{1}{2} \times 10^4 \times 3.17 = 1.585 \times 10^4\ m = 15\ km.$

3.   Calculate the mass of an a-particle. Its binding energy is $28\cdot2\ MeV.$

Let the mass of ‘$\alpha$’ particle be $xu$. ‘$\alpha$’ particle contains $2$ protons and $2$ neutrons. $\\$ $\therefore$ Binding energy $= (2 \times 1.007825\ u \times 1 \times 1.00866\ u\ –\ xu)C^2 = 28.2\ MeV$ (given). $\therefore x = 4.0016\ u.$

4.   How much energy is released in the following reaction? $$^7Li + p \rightarrow \alpha + \alpha$$ Atomic mass of $^7Li - 7\cdot0160\ u$ and that of $^4He - 4\cdot0026\ u$.

$Li^7 + p \rightarrow I + \alpha + E ;\ Li^7 = 7.016u$ $\\$ $\alpha = ^4He = 4.0026u ;\ p = 1.007276\ u$ $\\$ $E = Li^7 + P – 2\alpha = (7.016 + 1.007276)u – (2 \times 4.0026)u = 0.018076\ u$. $\\$ $\Rightarrow 0.018076 \times 931 = 16.828 = 16.83\ MeV.$

5.   Find the binding energy per nucleon of $_{79}^{97}Au$ if its atomic mass is $196\cdot96\ u.$

$B = (Zm_p + Nm_n – M)C^2$ $\\$ $Z = 79 ;\ N = 118 ;\ mp = 1.007276u ;\ M = 196.96\ u ;\ mn = 1.008665u$ $\\$ $B = [(79 \times 1.007276 + 118 \times 1.008665)u – Mu]c^2$ $\\$ $= 198.597274 \times 931 – 196.96 \times 931 = 1524.302094$ $\\$ so, Binding Energy per nucleon $= \dfrac{1524.3}{197} = 7.737.$

6.   (a) Calculate the energy released if $^{238}U$ emits an a-particle. (b) Calculate the energy to be supplied to $^{238}U$ if two protons and two neutrons are to be emitted one by one. The atomic masses of $^{238}U$, $^{234}Th$ and $^4He$ are $238\cdot0508\ u,\ 234\cdot04363\ u$ and $4\cdot00260\ u$ respectively.

a) $U^{238}\ _2He^4 + Th^{234}$ $\\$ $E = [M_u – (N_{HC} + M_{Th})]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487\ Mev = 4.255\ Mev$. b) $E = U^{238} – [Th^{234} + 2n'0 + 2p'1]$ $\\$ $= {238.0508 – [234.64363 + 2(1.008665) + 2(1.007276)]}u$ $\\$ $= 0.024712u = 23.0068 = 23.007\ MeV.$

7.   Find the energy liberated in the reaction $$^{223}Ra -\ ^{209}Pb +\ ^{14}C$$. $\\$ The atomic masses needed are as follows:$\\$ $$ ^{223}Ra \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{209}Pb \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{14}C$$ $\\$ $$223\cdot018\ u \ \ \ \ \ \ \ \ \ 208\cdot981\ u \ \ \ \ \ \ \ \ \ 14\cdot003\ u$$

$^{223}R_a = 223.018\ u ;\ ^{209}Pb = 208.981\ u ;\ ^{14}C = 14.003\ u$. $\\$ $^{223}R_a \rightarrow ^{209}Pb + ^{14}C$ $\\$ $\Delta{m} = mass\ ^{223}R_a\ –\ mass\ (^{209}Pb + ^{14}C)$ $\\$ $\Rightarrow = 223.018 – (208.981 + 14.003) = 0.034.$ $\\$ Energy $= \Delta{M} \times u = 0.034 \times 931 = 31.65 Me.$

8.   Show that the minimum energy needed to separate a proton from a nucleus with $Z$ protons and $N$ neutrons is $$\Delta{E}-(M_{Z - 1. N} + M_H - M_{Z.N})C^2$$ where $M_{Z. N} =$ mass of an atom with $Z$ protons and $N$ neutrons in the nucleus and $M_H =$ mass of a hydrogen atom. This energy is known as $proton-separation\ energy.$

$E_{Z.N}. \rightarrow E_{Z–1},\ N + P_1 \Rightarrow E_{Z.N}. \rightarrow E_{Z–1},\ N + _1H^1$ [As hydrogen has no neutrons but protons only] $\\$ $\Delta{E} = (M_{Z–1, N} + N_H – M_{Z,N})c^2$

9.   Calculate the minimum energy needed to separate a neutron from a nucleus with $Z$ protons and $N$ neutrons in terms of the masses $M_{Z. N}, M_{Z. N = 1},$ and the mass of the neutron.

$E_{2}N = E_{Z,N–1} + ^{1}_{0}n $. $\\$ Energy released = (Initial Mass of nucleus – Final mass of nucleus)$c^2 = (M_{Z.N –1} + M_0 – M_{ZN})c^2.$

10.   $^{32}P$ beta-decays to $^{32}S$. Find the sum of the energy of the antineutrino and the kinetic energy of the $\beta$ - particle. Neglect the recoil of the daughter nucleus. Atomic mass of $^{32}P = 31\cdot974\ u$ and that of $^{32}S = 31\cdot972\ u.$

$P^{32} \rightarrow S^{32}\ +\ _0v^{-0}\ +\ _1\beta^0$ $\\$ Energy of antineutrino and $\beta$ - particle $\\$ $= (31.974 – 31.972)u = 0.002\ u = 0.002 \times 931 = 1.862\ MeV = 1.86.$

11.   A free neutron beta-decays to a proton with a half-life of $14$ minutes, (a) What is the decay constant ? (b) Find the energy liberated in the process.

In $\rightarrow P + e^–$ $\\$ We know : Half life $= \dfrac{0.6931}{\lambda}$ (Where $\lambda =$ decay constant). $\\$ Or $\lambda = 0.6931 / 14 \times 60 = 8.25 \times 10{–4} S \ \ \ \ \ \ \ \ $ [As half life $= 14 min = 14 \times 60 sec$]. $\\$ Energy $= [M_n – (M_P + M_e)]u = [(M_{nu} – M_{pu}) – M_{pu}]c^2 = [0.00189u – 511\ KeV/c2] $ $\\$ $= [1293159\ ev/c^2 – 511000\ ev/c^2]c^2 = 782159\ eV = 782\ Kev$

12.   Complete the following decay schemes. $\\$ (a) $^{226}_{88}Ra \rightarrow\ \alpha\ +$ $\\$ (b) $^{19}_{8}O \rightarrow\ ^{19}_{9}F\ +$ $\\$ (c) $^{25}_{13}AI\ \rightarrow \ ^{25}_{12}MS\ +$

$^{226}_{58}Ra \rightarrow\ ^{4}_{2}\alpha +\ ^{222}_{26}Rn$ $\\$ $^{19}_{8}O \rightarrow\ ^{19}_{9}F +\ ^{0}_{n}\beta\ +\ ^{0-}_{0}v$ $\\$ $^{13}_{25}AI \rightarrow\ ^{25}_{12}MG +\ ^{0}_{-1}e +\ ^{0-}_{0}v$ $\\$

13.   In the decay $^{64}Cu \rightarrow ^{64}Ni + e^ * + v,$ the maximum kinetic energy carried by the positron is found to be $0\cdot650\ MeV.$ (a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy $0\cdot150\ MeV$ ? (b) What is the momentum of this neutrino in $kg-m/s$ ? Use the formula applicable to a photon.

$^{64}Cu \rightarrow\ ^{64}Ni + e^- + v$ $\\$ Emission of nutrino is along with a positron emission. $\\$ a) Energy of positron $= 0.650\ MeV$. $\\$ Energy of Nutrino $= 0.650 – KE$ of given position $= 0.650 – 0.150 = 0.5\ MeV = 500\ Kev$. $\\$ b) Momentum of Nutrino $= \dfrac{500 \times 1.6 \times 10^{-19}}{3 \times 10^8} \times 10^3 j = 2.67 \times 10 ^{22}\ kg m/s.$

14.   Potassium - $40$ can decay in three modes. It can decay by $\beta$ - emission, $\beta^*$ - emission or electron capture, (a) Write the equations showing the end products, (b) Find the $Q$-values in each of the three cases. Atomic masses of $^{40}_{18}Ar,\ ^{40}_{9}K$ and $^{11}_{20}Ca$ are $39\cdot9624\ u,\ 39\cdot9640\ u$ and $39\cdot9626\ u$ respectively.

a) $_{19}K^{40} \rightarrow\ _{20}Ca^{40} +\ _{-1}e^{0}\ +\ _{0}v^{0}$ $\\$ $_{19}K^{40} \rightarrow\ _{18}Ar^{40} +\ _{-1}e^{0} +\ _{0}v^{0}$ $\\$ $_{19}K^{40} +\ _{-1}e^{0} \rightarrow\ _{18}Ar^{40}$ $\\$ $_{19}K^{40} \rightarrow\ _{20}Ca^{40}\ +\ _{-1}e^{0} +\ _{0}v^{0} $.

b) $Q =$ [Mass of reactants – Mass of products]$c^2$ $\\$ $= [39.964u – 39.9626u] = [39.964u – 39.9626]uc^2 = (39.964 – 39.9626)\ 931\ Mev = 1.3034\ Mev$. $\\$ $_{19}K^{40} \rightarrow\ _{18}Ar^{40} + {-1}e^{0} + _{0}v^{0}$ $\\$ $Q = (39.9640 – 39.9624)uc^2 = 1.4890 = 1.49 Mev.$ $\\$ $_{19}K^{40} + {-1}e^{0} \rightarrow\ _{18}Ar^{40}$ $\\$ $Q_{value} = (39.964 – 39.9624)uc^2$.

15.   Lithium $(Z = 3)$ has two stable isotopes $^6Li$ and $^7Li.$ When neutrons are bombarded on lithium sample, electrons and $\alpha$ -particles are ejected. Write down the nuclear processes taking place.

$_{6}^{3}Li + n \rightarrow\ _{7}^{3}Li ;\ _{7}^{3}Li + r \rightarrow\ _{8}^{3}Li$ $\\$ $_{8}^{3}Li \rightarrow\ _{8}^{4}Be + e^- + v^-$ $\\$ $_{4}^{8}Be \rightarrow\ _{2}^{4}He +\ _{2}^{4}He$

16.   The masses of $^{11}C$ and $^{11}B$ are respectively $11\cdot0114\ u$ and $11\cdot0093\ u$. Find the maximum energy a positron can have in the $\beta^*$- decay of $^{11}C$ to $^{11}B$.

$"C \rightarrow "B + \beta^+ + v$ $\\$ mass of $C" = 11.014u $; mass of $B" = 11.0093u$ $\\$ Energy liberated $= (11.014 – 11.0093)u = 29.5127\ Mev.$ $\\$ For maximum $K.E.$ of the positron energy of v may be assumed as $0.$ $\\$ $\therefore$ Maximum $K.E.$ of the positron is $29.5127\ Mev.$

17.   $^{228}Th$ emits an alpha particle to reduce to $^{224}Ra$. Calculate the kinetic energy of the alpha particle emitted in the following decay: $$^{228}Th\ \rightarrow\ ^{224}Ra^ * + \alpha$$ $$^{224}Ra^*\ \rightarrow\ ^{224}Ra + \gamma\ (217\ keV).$$

18.   Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme: $$^{12}N \rightarrow\ ^{12}C^* + e^* + v $$ $$^{12}C^* \rightarrow\ ^{I2}C + \gamma\ (4\cdot43 MeV)$$. The atomic mass of $^{12}N$ is $12\cdot018613\ u.$

Mass $^{238}Th = 228.028726\ u ;\ ^{224}Ra = 224.020196\ u ;\ \alpha = ^{4} _{2}He \rightarrow 4.00260u$ $\\$ $^{238}Th \rightarrow ^{224}Ra* + \alpha$ $\\$ $^{224}Ra^* \rightarrow ^{224}Ra + v(217\ Kev)$ $\\$ Now, Mass of $^{224}Ra* = 224.020196 \times 931 + 0.217\ Mev = 208563.0195\ Mev.$ $\\$ $KE of \alpha = E ^{226}Th – E(^{224}Ra* + \alpha)$ $\\$ $= 228.028726 \times 931 – [208563.0195 + 4.00260 \times 931] = 5.30383\ Mev = 5.304\ Mev.$

19.   The decay constant of $^{197}_{60}Hg$ (electron capture to $^{197}_{9}Au$) is $1\cdot8 \times 10^{-4}\ s^{-1}$ (a) What is the half-life ? (b) What is the average-life ? (c) How much time will it take to convert $25$% of this isotope of mercury into gold ?

$^{12}N \rightarrow ^{12}C* + e^+ + v$ $\\$ $^{12}C* \rightarrow ^{12}C + v(4.43\ Mev)$ $\\$ Net reaction $: ^{12}N \rightarrow ^{12}C + e^+ + v + v(4.43\ Mev)$ $\\$ Energy of $(e^+ + v) = N^{12} – (c^{12} + v)$ $\\$ $= 12.018613u – (12)u – 4.43 = 0.018613\ u – 4.43 = 17.328 – 4.43 = 12.89\ Mev.$ $\\$ Maximum energy of electron (assuming $0$ energy for $v$) $= 12.89\ Mev.$

20.   The half-life of $^{198}Au$ is $2\cdot7$ days, (a) Find the activity of a sample containing $1\cdot0\ \mu{g}$ of $^{198}Au.$ (b) What will be the activity after $7$ days ? Take the atomic weight of $^{198}Au$ to be $198\ g/mol.$

a) $t_{1/2} = \dfrac{0.693}{\lambda}$ [$\lambda \rightarrow$ Decay constant] $\Rightarrow t_{1/2} = 3820$ sec $= 64$ min. $\\$ b) Average life $= \dfrac{t_{1/2}} {0.693} = 92$ min. $\\$ c) $0.75 = 1\ e^{–\lambda{t}} \Rightarrow$ In $0.75 = – \lambda{t} \Rightarrow t =$ In $\dfrac{0.75}{– 0.00018} = 1598.23\ sec.$ $\\$

21.   The half-life of $^{198}Au$ is $2\cdot7$ days, (a) Find the activity of a sample containing $1\cdot00\ \mu{g}$ of $^{198}Au$. (b) What will be the activity after $7$ days ? Take the atomic weight of $^{198}Au$ to be $198\ g/mol.$

a) $198$ grams of $Ag$ contains $\rightarrow N_0$ atoms. $\\$ $1 \mu{g}$ of $Ag$ contains $\rightarrow \dfrac{N_{0}}{198} \times 1 \mu{g} = \dfrac{6 \times 10^{23} \times 1 \times 10^6}{198}$ atoms $\\$

Activity $= \lambda{N} = \dfrac{0.963}{t_{1/ 2}} \times N = \dfrac{0.693 \times 6 \times 10^{17}}{198 \times 2.7}$ disintegrations/day. $\\$ $= \dfrac{0.693 \times 6 \times 10^{17}} {198 \times 2.7 \times 3600 \times 24}$ disintegration/sec $= \dfrac{0.693 \times 6 \times 10^{17}}{198 \times 2.7 \times 36 \times 24 \times 3.7 \times 10^{10}}$curie $= 0.244$ Curie

22.   Radioactive $^{131}1$ has a half-life of $8\cdot0 $ days. A sample containing 1311 has activity $20\ \mu Ci$ at $t = 0$. (a) What is its activity at $t = 4\cdot0$ days ? (b) What is its decay constant at $t - 4\cdot0$ days ?

$t_{1}{2} = 8.0\ days ; A_0 = 20 \mu\ Cl$ $\\$ a) $t = 4.0\ days ; \lambda = \dfrac{0.693}{8}$ $\\$ $ A = A0e^{–\lambda{t}} = 20 \times 10^{–6} \times e^{(-0.693 / 8) \times 4} = 1.41 \times 10^{–5}\ Ci = 14\ \mu\ Ci$ $\\$ b) $\lambda = {0.693}{8 \times 24 \times 3600} = 1.0026 \times 10^{–6}.$ $\\$

23.   The decay constant of $^{238}U$ is $4\cdot9 \times 10^{18}\ s^{-1}$. (a) What is the average-life of $^{238}U$ ? (b) What is the half-life of $^{238}U$ ? (c) By what factor does the activity of a $^{238}U$ sample decrease in $9 \times 10^9$ years ?

$\lambda = 4.9 \times 10^{–18}\ s^{–1}$ $\\$ a) Avg. life of $^{238}U = \dfrac{1}{\lambda}= \dfrac{1}{4.9 \times 10^{-18}} = \dfrac{1}{4.9} \times 10^{-18}$ sec. $\\$ $= 6.47 \times 10^3$ years. $\\$ b) Half life of uranium $= \dfrac{0.693}{\lambda} = \dfrac{0.693}{4.9 \times 10^{-18}} = 4.5 \times 10^9$ years. $\\$ c) $A = \dfrac{A_0}{2^{t / t_{1/ 2}}} \Rightarrow {A_0}{A} = 2^{t/t_{1/2}} = 2^2 = 4.$

24.   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, (a) What is the decay constant of the sample ? (b) What is its half-life ?

$A = 200,\ A_0 = 500,\ t = 50\ min$ $\\$ $A = A_0 e^{–\lambda{t}}$ or $200 = 500 \times e^{–50 \times 60 \times \lambda}$ $\\$ $\Rightarrow \lambda = 3.05 \times 10^{–4}\ s.$ $\\$ b) $t_{1/2} = \dfrac{0.693}{\lambda} = \dfrac{0.693}{0.000305}= 2272.13\ sec = 38\ min.$

25.   The count rate from a radioactive sample falls from $4\cdot0 \times 10^6$ per second to $1\cdot0 \times 10^6$ per second in $20$ hours. What will be the count rate $100$ hours after the beginning ?

Answer

25   None

$A_0 = 4 \times 10^5\ disintegration / sec$ $\\$ $A' = 1 \times 10^6\ dis/sec ;\ t = 20\ hours.$ $\\$ $A' = \dfrac{A_0}{2^{t / t_{1/ 2}}} \Rightarrow 2^{t / t_{1/ 2}} = \dfrac{A_0}{A'} \Rightarrow 2^{t / t_{1/ 2}} = 4$ $\\$ $\Rightarrow t / t_{1/ 2} = 2 \Rightarrow t_{1/2} = \dfrac{t}{2} = \dfrac{20\ hours}{2} = 10\ hours.$ $\\$ $A" = \dfrac{A_0}{2^{t / t _{1/ 2}}} \Rightarrow A" = \Rightarrow \dfrac{4 \times 10^6}{2^{100 /10}} = 0.00390625 \times 10^6 = 3.9 \times 10^3$ dintegrations/sec.

26.   The half-life of $^{220}Ra$ is $1602\ y$. Calculate the activity of $0\cdot1\ g$ of $RaCl_2$ in which all the radium is in the form of $^{220}Ra$. Taken atomic weight of $Ra$ to be $226\ g/mol$ and that of $CI$ to be $35\cdot5\ g/mol.$

Answer

26   None

$t_{1/2} = 1602\ Y ;\ Ra = 226\ g/mole ;\ Cl = 35.5\ g/mole.$ $\\$ $1\ mole\ RaCl_2 = 226 + 71 = 297\ g$ $\\$ $297g = 1 mole of Ra.$ $\\$ $0.1\ g = \dfrac{1}{297} \times 0.1 mole of Ra = \dfrac{0.1 \times 6.023 \times 10^{23}}{297} = 0.02027 \times 10^{22}$ $\\$ $\lambda = \dfrac{0.693}{t_{1/2}} = 1.371 \times 10{–11}.$ $\\$ Activity $= \lambda{N} = 1.371 \times 10^{–11} \times 2.027 \times 10^{20} = 2.779 \times 10^9 = 2.8 \times 10^9$ disintegrations/second

27.   The half-life of a radioisotope is $10\ h$. Find the total number of disintegrations in the tenth hour measured from a time when the activity was $1\ Ci.$

$t_{1/2} = 10$ hours, $A_0 = 1\ ci$ $\\$ Activity after $9$ hours $= A_0\ e–^{\lambda{t}} = 1 \times e \dfrac{0.693}{10} \times 9 = 0.5359 = 0.536\ Ci.$ $\\$ No. of atoms left after $9^{th}$ hour, $A_9 = \lambda N_9$ $\\$ $\Rightarrow N_9 = {A_9}{\lambda} = \dfrac{0.536 \times 10 \times 3.7 \times 10^{10} \times 3600}{0.693}= 28.6176 \times 10^{10} \times 3600 = 103.023 \times 10^{13}.$

Activity after $10$ hours $= A_0\ e^{–\lambda{t}} = 1 \times e \dfrac{-0.693}{10} \times 9 = 0.5\ Ci.$ $\\$ No. of atoms left after $10^th$ hour $\\$ $A_10 = \lambda N_10$

$\Rightarrow N_{10} =A_{10} = \dfrac{0.5 \times 3.7 \times 10^{10} \times 3600}{0.693 /10} = 26.37 \times 10^{10} \times 3600 = 96.103 \times 10^{13}$. $\\$ No.of disintegrations $= (103.023 – 96.103) \times 10^{13} = 6.92 \times 10^{13}.$

28.   The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a one month old $^{32}P$($t_{1/2} = 14\cdot3$ days) source if it was originally purchased for $800$ rupees ?

Answer

28   None

$t_{1/2} = 14.3\ days ;\ t = 30\ days = 1 month$ $\\$ As, the selling rate is decided by the activity, hence $A_0 = 800$ disintegration/sec. $\\$ We know, $A = A_{0}e^{–\lambda{t}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[\lambda = \dfrac{0.693}{14.3}\Big]$ $\\$ $A = 800 \times 0.233669 = 186.935 = 187$ rupees.

29.   $^{57}Co$ decays to $^{57}Fe$ by $\beta$ - emission. The resulting $^{57}Fe$ is in its excited state and comes to the ground state by emitting $y$-rays. The half-life of $\beta$ - decay is $270$ days and that of the y-emission is $10^8$ s. A sample of $^{57}Co$ gives $5\cdot0 \times 10^5$ gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to $2\cdot5 \times 10^9$ per second ?

Answer

29   None

According to the question, the emission rate of $\gamma$ rays will drop to half when the $\beta^+$ decays to half of its original amount. And for this the sample would take $270$ days. $\\$ $\therefore$ The required time is $270$ days.

30.   Carbon $(Z = 6)$ with mass number $11$ decays to boron $(Z = 5)$. (a) Is it a $\beta^+$ - decay or a $\beta^-$ - decay ? (b) The half-life of the decay scheme is $20\cdot3$ 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

30   None

a)$P \rightarrow n + e^+ + v$ Hence it is a $\beta^+$ decay. $\\$ b) Let the total no. of atoms be $100\ N_0.$ $\\$ $ \ \ \ $ Carbon $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ Boron $\\$ Initially $90\ N_0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 10\ N_0$ $\\$ Finally $10\ N_0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 90\ N_0$

Now, $10\ N_0 = 90\ N_0\ e^{–\lambda{t}} \Rightarrow \dfrac{1}{9} = e^{\dfrac{-0.693}{20.3}} \times t$ $\\$ [because $t_{1/2} = 20.3$ min] $\\$ $\Rightarrow In \dfrac{1}{9} = \dfrac{-0.693}{20.3}t \Rightarrow t = \dfrac{2.1972 \times 20.3}{0.693} = 64.36 = 64\ min.$

31.   $4 \times 10^{23}$ tritium atoms are contained in a vessel. The half-life of decay of tritium nuclei is $12\cdot3\ y.$ Find (a) the activity of the sample, (b) the number of decays in the next $10$ hours (c) the number of decays in the next $6\cdot15\ y.$

Answer

31   None

$N = 4 \times 10^{23} ;\ t_{1/2} = 12.3$ years. $\\$ a) Activity $= \dfrac{dN}{dt} = \lambda{n} = \dfrac{0.693}{t_{1/ 2}}N = {0.693}{12.3} \times 4 \times 10^{23}$ dis/year

$= 7.146 \times 10^{14}$ dis/sec. $\\$\ b) $\dfrac{dN}{dt} = 7.146 \times 10^{14}$ $\\$ No.of decays in next $10$ hours $= 7.146 \times 10{14} \times 10 \times 36.. = 257.256 \times 10^{17} = 2.57 \times 10^{19}.$ $\\$ c) $N = N_0\ e^{–\lambda{t}} = 4 \times 10^{23} \times e^{\frac{0.693}{20.3}} \times 6.16 = 2.82 \times 10^{23} =$ No.of atoms remained $\\$ No. of atoms disintegrated $= (4 – 2.82) \times 10^{23} = 1.18 \times 10^{23}.$

32.   A point source emitting alpha particles is placed at a distance of $1\ m$ from a counter which records any alpha particle falling on its $1\ cm^2$ window. If the source contains $6\cdot0 \times 10$ active nuclei and the counter records a rate of $50000$ counts/second, find the decay constant. Assume that the source emits alpha particles uniformly in all directions and the alpha particles fall nearly normally on the window.

Answer

32   None

Counts received per $cm^2 = 50000\ Counts/sec.$ $\\$ $N = N_{3}o$ of active nucleic $= 6 \times 10^{16}$ $\\$ Total counts radiated from the source = Total surface area $\times 50000\ counts/cm^2$ $\\$ $= 4 \times 3.14 \times 1 \times 10^4 \times 5 \times 10^4 = 6.28 \times 10^9$ Counts $= \dfrac{dN}{dt}$ $\\$ We know, $\dfrac{dN}{dt} = \lambda{N}$ $\\$ Or $\lambda = \dfrac{6.28 \times 10^9}{6 \times 10^{16}} = 1.0467 \times 10^{–7} = 1.05 \times 10^{–7}\ s^{–1}.$

33.   $^{238}U$ decays to $^{200}Pb$ with a half-life of $4\cdot47 \times 10^9\ y$. This happens in a number of steps. Can you justify a single half-life for this chain of processes ? A sample of rock is found to contain $2\cdot00\ mg$ of $^{238}U$ and $0\cdot600\ mg$ of $^{200}Pb$. Assuming that all the lead has come from uranium, find the life of the rock.

Answer

33   None

Half life period can be a single for all the process. It is the time taken for $1/2$ of the uranium to convert to lead. $\\$ No. of atoms of $U^{238} =\dfrac{6 \times 10^{23} \times 2 \times 10^{-3}}{238} = \dfrac{12}{238} \times 10^{20} = 0.05042 \times 10^{20}$

No. of atoms in $Pb = \dfrac{6 \times 10^{23} \times 0.6 \times 106{-3}}{206} = \dfrac{3.6}{206} \times 10^{20}$

Initially total no. of uranium atoms $= \dfrac{12}{235} + \dfrac{3.6}{206}\Big) \times 10^{20} = 0.06789$ $\\$ $N = N_0\ e^{–\lambda{t}} \Rightarrow N = N_0\ e^{\frac{-0.693}{t / t_{1/ 2}}} \Rightarrow 0.05042 = 0.06789\ e^{\frac{0.693}{4.47 \times 10^9}}$ $\\$ $\Rightarrow log \Big(\dfrac{0.05042}{0.06789}\Big) = \dfrac{0.693t}{4.47 \times 10^9}$ $\\$ $\Rightarrow t = 1.92 \times 10^9\ years.$

34.   When charcoal is prepared from a living tree, it shows a disintegration rate of $15\cdot3$ disintegrations of $^{14}C$ per gram per minute. A sample from an ancient piece of charcoal shows HC activity to be $12\cdot3$ disintegrations per gram per minute. How old is this sample ? Half-life of $^{14}C$ is $5730\ y.$

Answer

34   None

$A_0 = 15.3 ;\ A = 12.3 ;\ t_{1/2} = 5730\ year$ $\\$ $\lambda = \dfrac{0.6931}{T_{1/ 2}} \dfrac{0.6931}{5730} yr^{-1}$ $\\$ Let the time passed be $t$, We know $A = A_0\ e^{-\lambda{t}} - \dfrac{0.6931}{5730} \times t \Rightarrow 12.3 = 15.3 \times e$. $\\$ $\Rightarrow t = 1804.3\ years.$

35.   Natural water contains a small amount of tritium $(^3H)$. This isotope beta-decays with a half-life of $12\cdot5$ years. A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things he finds a sealed bottle of whisky. On return he analyses the whisky and finds that it contains only $1\cdot5$ per cent of the $^3H$ radioactivity as compared to a recently purchased bottle marked '$8$ years old'. Estimate the time of that unsuccessful attempt.

Answer

35   None

The activity when the bottle was manufactured $= A_0$ Activity after $8$ years $= A_0e^{\frac{12.5}{0.693}} \times 8$ $\\$ Let the time of the mountaineering = t years from the present $\\$ $A = A_0e^{\frac{0.693}{12.5}} \times t ; A =$ Activity of the bottle found on the mountain. $\\$ A = (Activity of the bottle manufactured 8 years before) \times $1.5$% $\Rightarrow A_0e^{\frac{0.693}{12.5}} = A_0e^{\frac{12.5}{0.693}} \times 8 \times 0.015 $ $\\$ $\dfrac{0.693}{12.5}t = \dfrac{0.693 \times 8}{12.5} + In[0.015]$ $\\$ $\Rightarrow 0.05544\ t = 0.44352 + 4.1997 \Rightarrow t = 83.75\ years.$

36.   The count rate of nuclear radiation coming from a radioactive sample containing $^{128}I$ varies with time as follows. $\\$ Time $t$ (minute) $: \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ 25 \ \ \ \ 50 \ \ \ \ \ \ \ 75 \ \ \ \ \ \ 100$ $\\$ Count rate $R(10^9\ s^{-1})$ $: \ \ \ 30 \ \ \ 16 \ \ \ \ 8\cdot0 \ \ \ \ 3\cdot8 \ \ \ \ \ 2 0$ $\\$ (a) Plot $ln(R_0/R)$ against $t$. (b) From the slope of. the best straight line through the points, find the decay constant \lambda. (c) Calculate the half-life $T_{\frac{1}{2}}$.

Answer

36   None

a) Here we should take $R_0$ at time is $t_0 = 30 \times 10^9\ s^{–1}$ $\\$ i) $In(R_0/R_1) = In \Big(\dfrac{30 \times 10^9}{30 \times 10^9}\Big) = 0$ $\\$ ii) $In(R_0/R_2) = In \Big(\dfrac{30 \times 10^9}{16 \times 10^9}\Big) = 0.634$ $\\$ iii) $In(R_0/R_3) = In \Big(\dfrac{30 \times 10^9}{8 \times 10^9}\Big) = 1.35$ $\\$ iv) $In(R_0/R_4) = In \Big(\dfrac{30 \times 10^9}{3.8 \times 10^9}\Big) = 2.06$ $\\$ i) $In(R_0/R_5) = In \Big(\dfrac{30 \times 10^9}{2 \times 10^9}\Big) = 2.7$

b) $\therefore$ The decay constant $\lambda = 0.028\ min^{–1}$ $\\$ c) $\therefore$ The half life period $= t_{1/2}$. $\\$ $t_{1/2} = \dfrac{0.693}{\lambda} = \dfrac{0.693}{0.028} = 25\ min.$

37.   The half-life of $^{40}K$ is $1\cdot30 \times 10^9 y$. A sample of $1\cdot00 g$ of pure $KCI$ gives $160$ country's. Calculate the relative abundance of $^{40}K$ (fraction of $^{40}K$ present) in natural potassium.

Answer

37   None

Given : Half life period $t_{1/2} = 1.30 \times 10^9\ year ,\ A = 160\ count/s = 1.30 \times 10^9 \times 365 \times 86400$ $\\$ $\therefore A = \lambda{N} \Rightarrow 160 = \dfrac{0.693}{t_{1/ 2}}N$ $\\$ $\Rightarrow N = \dfrac{160 \times 1.30 \times 365 \times 86400 \times 10^9}{0.693}= 9.5 \times 10^{18}$ $\\$ $\therefore 6.023 \times 10^{23}$ No. of present in $40$ grams.$\\$ $6.023 \times 10^{23} = 40\ g \Rightarrow 1 = \dfrac{40}{6.023 \times 10^{23}}$ $\\$ $\therefore 9.5 \times 10^{18}$ present in $ = \dfrac{40 \times 9.5 \times 10^{18}}{6.023 \times 10^{23}} = 6.309 \times 10^{–4} = 0.00063.$ $\\$ $\therefore$ The relative abundance at $40\ k$ in natural potassium $= (2 \times 0.00063 \times 100)$% $= 0.12$%.$

38.   $^{197}_{80}Hg$ decays to $^{197}_{80}Au$ through electron capture with a decay constant of $0\cdot257$ per day. (a) What other particle or particles are emitted in the decay ? (b) Assume that the electron is captured from the $K$ shell. Use Moseley's law $\sqrt{v} = a(Z - b)$ with $a = 4\cdot95 \times 10^7\ s ^{-1/2}$ and $b = 1$ to find the wavelength of the $K_a\ X$-ray emitted following the electron capture.

Answer

38   None

a) $P + e \rightarrow n + v$ neutrino $[a \rightarrow 4.95 \times 10^7\ s^{-1/2} ;\ b \rightarrow 1]$ $\\$ b) $\sqrt{f} = a(z – b)$ $\\$ $\Rightarrow \sqrt{c / \lambda} = 4.95 \times 107\ (79 – 1) = 4.95 \times 10^7 \times 78 \Rightarrow C/\lambda = (4.95 \times 78)^2 \times 10^{14}$ $\\$ $\Rightarrow \lambda = \dfrac{3 \times 10^{8}}{14903.2 \times 10^{14}}= 2 \times 10^{–5} \times 10^{–6} = 2 \times 10^{–4}\ m = 20\ pm.$

39.   A radioactive isotope is being produced at a constant rate $dN/dt = R$ in an experiment. The isotope has a half-life $t_{1/2}$. Show that after a time $t >> t_{1/2}$, the number of active nuclei will become constant. Find the value of this constant.

Answer

39   None

Given : Half life period $= t_{1/2}$, Rate of radio active decay $= \dfrac{dN}{dt} = R \Rightarrow R = \dfrac{dN}{dt}$ $\\$ Given after time $t >> t_{1/2}$, the number of active nuclei will become constant. $\\$ i.e. $\Big(\dfrac{dN}{dt}\Big)_{present} = R = \Big(\dfrac{dN}{dt}\Big)_{decay}$ $\\$

$\therefore R = \dfrac{dN}{dt}_{decay}$ $\\$ $\Rightarrow R = \lambda{N}$ [where, $\lambda =$ Radioactive decay constant, $N =$ constant number] $\\$ $\Rightarrow R = {0.693}{t_{1/ 2}}(N) \Rightarrow Rt_{1/2} = 0.693\ N \Rightarrow N = \dfrac{Rt_{1/ 2}}{0.693}$

40.   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 $t.$

Answer

40   None

Let $N_0 =$ No. of radioactive particle present at time $t = 0$ $\\$ $N =$ No. of radio active particle present at time $t.$ $\therefore N = N_0\ e^{–\lambda{t}}$ $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ [$\lambda -$ Radioactive decay constant] $\\$ $\therefore$ The no.of particles decay $= N_0 – N = N_0 – N_0e^{–\lambda{t}} = N_0\ (1 – e^{–\lambda{t}})$ $\\$ We know, $A_0 = \lambda N_0 ;\ R = \lambda N_0 ; N_0 = R/\lambda$ $\\$ From the above equation $\\$ $N = N_0\ (1 – e^{–\lambda{t}}) = \dfrac{R}{\lambda} (1- e^{-\lambda{t}} ) \ \ \ \ \ \ \ \ \ \ \ $ (substituting the value of $N_0$)

41.   In an agricultural experiment, a solution containing $1$ mole of a radioactive material ($t_{1/2} = 14\cdot3$ days) was injected into the ruots of a plant. The plant was allowed $70$ hours to settle down and then activity was measured in its fruit. If the activity measured was $1\ \mu Ci$, what per cent of activity is transmitted from the root to the fruit in steady state ?

Answer

41   None

$n = 1\ mole = 6 \times 10^{23}\ atoms,\ t_{1/2} = 14.3\ days$ $\\$ $t = 70\ hours,\ \dfrac{dN}{dt}$ in root after time $t = \lambda{N}$ $\\$

$N = N_O\ e^{–\lambda{t}} = 6 \times 10^{23} \times e^\frac{-0.693 \times 70}{14.3 \times 24} = 6 \times 10^{23} \times 0.868 = 5.209 \times 10^{23}$. $\\$ $0.0105 \times 10^{23} \times \dfrac{-0.693}{14.3 \times 24} = \dfrac{5.209 \times 10^{23}}{3600}\ dis/hour.$ $\\$ $= 2.9 \times 10^{–6} \times 10^{23}\ dis/sec = 2.9 \times 10^{17}\ dis/sec.$ $\\$ Fraction of activity transmitted $= \dfrac{1\mu ci}{2.9 \times 10^{17}} \times 100$% $\\$ $\Rightarrow \Big(\dfrac{1 \times 3.7 \times 10^{8}}{2.9 \times 10^{11}} \times 100\Big)% = 1.275 \times 10^{–11}$ %.

42.   A vessel of volume $125\ cm^3$ contains tritium $t_{1/2}= 12\cdot3\ y)$ at $500\ kPa$ and $300\ K$. Calculate the activity of the gas.

Answer

42   None

$V = 125\ cm^3 = 0.125\ L,\ P = 500\ K\ pa = 5\ atm$. $T = 300\ K, t_{1/2} = 12.3\ years = 3.82 \times 10^{8}\ sec.$ Activity $= \lambda \times N$ $\\$ $N = n \times 6.023 \times 10^{23} = \dfrac{5 \times 0.125}{8.2 \times 10^{-2} \times 3 \times 10^{2}} \times 6.023 \times 10^{23} = 1.5 \times 10^{22} atoms.$ $\\$ $\lambda = \dfrac{0.693}{3.82 \times 10^{-8}} = 0.1814 \times 10^{–8} = 1.81 \times 10^{–9}\ s^{–1}$ $\\$ Activity $= \lambda{N} = 1.81 \times 10^{–9} \times 1.5 \times 10^{22} = 2.7 \times 10^{3}\ disintegration/sec$ $\\$ $= \dfrac{2.7 \times 10^{13}}{3.7 \times 10^{10}}Ci = 729\ Ci.$

43.   $^{212}_{83}Bi$ can disintegrate either by emitting an $\alpha$-particle or by emitting a $\beta^-$-particle, (a) Write the two equations showing the products of the decays, (b) The probabilities of disintegration by $\alpha$- and $\beta$-decays are in the ratio $7/13$. The overall half-life of $^{212}Bi$ is one hour. If $1\ g$ of pure $^{212}Bi$ is taken at $12\cdot00\ noon$, what will be the composition of this sample at $1\ p.m$. the same day ?

$^{212}_{83}Bi \rightarrow ^{208}_{81}Ti + ^{4}_{2}He(\alpha)$ $\\$ $^{212}_{83}Bi \rightarrow ^{212}_{84}Bi \rightarrow ^{212}_{84}P_0 + e^-$ $\\$ $t_{1/2} = 1\ h. \ \ \ \ \ \ \ \ \ \ Time\ elapsed = 1\ hour$ $\\$ $at t = 0 Bi^{212} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Present = 1\ g$ $\\$ $\therefore at t = 1\ Bi^{212} \ \ \ \ \ \ \ \ \ \ \ \ \ Present = 0.5\ g$ $\\$ Probability $\alpha$-decay and $\beta$-decay are in ratio $7/13$. $\\$ $\therefore T_l$ remained $= 0.175\ g$ $\\$ $\therefore P_0$ remained $= 0.325\ g$ $\\$

44.   A sample contains a mixture of $^{106}Ag$ and $^{110}Ag$ isotopes each having an activity of $8\cdot0 \times10^{8}$ disintegrations per' second.$^{110}Ag$ is known to have larger half-life than $^{108}Ag$. The activity $A$ is measured as a function of time and the following data are obtained. $\\$ $Time\ (s) \ \ \ \ Activity (A) (10 8 disinte-grations/s) \ \ \ \ \ Time (s) \ \ \ \ Activity (A). (10 8 disinte-grations/s)$ $\\$ $20 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 11\cdot799 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\cdot0828$ $\\$ $40 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9\cdot1680 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 300 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\cdot8899$ $\\$ $60 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7\cdot4492 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 400 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\cdot1671$ $\\$ $80 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\cdot2684 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\cdot7212$ $\\$ $100 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 54115$ $\\$ (a) Plot $ln(A/A_0)$ versus time, (b) See that for large values of time, the plot is nearly linear. Deduce the half-life of $^{110}Ag$ from this portion of the plot, (c) Use the half-life of $^{110}Ag$ to calculate the activity corresponding to $^{108}Ag$ in the first $50\ s$. (d) Plot $ln(A/A_0)$ versus time for $^{108}Ag$ for the first $50\ s$. (e) Find the half-life of $^{I08}Ag$.

Activities of sample containing $^{108}Ag$ and $^{110}Ag$ isotopes $= 8.0 \times 10^{8}$ disintegration/sec.$\\$ a) Here we take $A = 8 \times 10^{8}$ dis./sec $\\$ $\therefore$ i) In $\Big(\dfrac{A_1}{A_{0_1}}\Big) = In \Big(\dfrac{11.794}{8}\Big) = 0.389$ $\\$ ii) In $\Big(\dfrac{A_2}{A_{0_2}}\Big) \ \ \ \ \ \ \ \ \ = In \Big(\dfrac{9.1680}{8}\Big) = 0.1362$ $\\$ iii) In $\Big(\dfrac{A_3}{A_{0_3}}\Big) \ \ \ \ \ \ \ \ \ = In \Big(\dfrac{7.4492}{8}\Big) = –0.072$ $\\$ iv) In $\Big(\dfrac{A_4}{A_{0_4}}\Big) \ \ \ \ \ \ \ \ \ = In \Big(\dfrac{6.2684}{8}\Big) = –0.244$ $\\$ v) In $\Big(\dfrac{5.4115}{8}\Big) = –0.391$ $\\$ vi) In $\Big(\dfrac{3.0828}{8}\Big) = –0.954$ $\\$ vii) In $\Big(\dfrac{1.8899}{8}\Big) = –1.443$ $\\$ viii) In $\Big(\dfrac{1.167}{8}\Big) = –1.93$ $\\$ ix) In $\Big(\dfrac{0.7212}{8}\Big) = –2.406$

b) The half life of $110\ Ag$ from this part of the plot is $24.4\ s$. $\\$ c) Half life of $^{110}Ag = 24.4\ s$. $\\$ $\therefore$ decay constant $\lambda = \dfrac{0.693}{24.4} = 0.0284 \Rightarrow t = 50\ sec,$ $\\$ The activity $A = A_0e{–\lambda{t}} = 8 \times 10^8 \times e^{–0.0284 \times 50} = 1.93 \times 10^8$

d) diagram $\\$ e) The half life period of $^{108}Ag$ from the graph is $144\ s$.

45.   A human body excretes (removes by waste discharge, sweating etc.) certain materials by a law similar to radioactivity. If technitium 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 $6\ \mu Ci$. How much time will elapse before the activity falls to $3\ \mu Ci$ ?

$t_{1/2} = 24\ h$ $\\$ $\therefore t_{1/2} = \dfrac{t_1 t_2}{t_1 + t_2} = \dfrac{24 \times 6}{24 + 6}= 4.8\ h.$ $\\$ $A_0 = 6\ rci ;\ A = 3\ rci$ $\\$ $\therefore A = \dfrac{A_0}{2^{t / t_{1/ 2}}}\Rightarrow 3\ rci = \dfrac{6\ rci}{2^{t / 4.8h}} \Rightarrow \dfrac{t}{24.8h} = 2 \Rightarrow t = 4.8\ h.$

46.   A charged capacitor of capacitance $C$ is discharged through a resistance $R$. $A$ radioactive sample decays with an average-life $t$. 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.

$Q = qe^{-t/CR} ;\ A = A_0e{–\lambda{t}}$ $\\$ $\dfrac{Energy}{Activity} = \dfrac {1q^2 \times e^{2t/cR}}{2 CA_0 e{-\lambda{t}}}$ $\\$ Since the term is independent of time, so their coefficients can be equated, $\\$ So, $\dfrac{2t}{CR} = \lambda{t} \ \ \ \ \ \ \ \ \ \ \ \ or, \lambda = \dfrac{2}{CR} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ or, = \dfrac{1}{\tau} = \dfrac{2}{CR} \ \ \ \ \ \ \ \ \ \ \ \ \ \ or, R = 2 \dfrac{\tau}{C}$ (Proved)

47.   Radioactive isotopes are produced in a nuclear physics experiment at a constant rate $\dfrac{dN}{dt} = R$. An inductor of inductance $100\ mH$, 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 $i/N$ 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.

$R = 100\ \Omega ;\ L = 100\ mH$ $\\$ After time $t, i = i_0 (1 - e^{-t /Lr}) \ \ \ \ \ \ \ \ \ \ \ N = N_0\ (e^{–\lambda{t}})$ $\\$ $ \dfrac{i}{N} = \dfrac{i_0(1 - e^{tR /L})}{N_0e^{-\lambda{t}}}\ i/N$ is constant i.e. independent of time. $\\$ Coefficients of t are equal $–R/L = –\lambda \Rightarrow R/L = \dfrac{0.693}{t_{1/2}}$ $\\$ $= t_{1/2} = 0.693 \times 10^{–3} = 6.93 \times 10^{–4} sec.$

48.   Calculate the energy released by $1\ g$ of natural uranium assuming $200\ MeV$ is released in each fission event and that the fissionable isotope $^{235}U$ has an abundance of $0\cdot7$% by weight in natural uranium.

$1\ g$ of $‘I’$ contain $0.007\ g\ U^{235} \ \ \ \ \ \ \ \ $ So, $235\ g$ contains $6.023 \times 10^{23}\ atoms.$ $\\$ So, $0.7\ g$ contains $\dfrac{6.023 \times 10^{23}}{235} \times 0.007\ atom$ $\\$ $1$ atom given $200\ Mev$. So, $0.7\ g$ contains $\dfrac{6.023 \times 10^{23} \times 0.007 \times 200 \times 10^6 \times 1.6 \times 10^{-19}}{235}J = 5.74 \times10^{–8}\ J.$ $\\$

49.   A uranium reactor develops thermal energy at a rate of $300\ MW$. Calculate the amount of $^{235}U$ being consumed every second. Average energy released per fission is $200\ MeV.$

Let $n$ atoms disintegrate per second Total energy emitted/sec $= (n \times 200 \times 10^6 \times 1.6 \times 10^{–19}) J$ = Power $\\$ $300$ MW $= 300 \times 10^6$ Watt = Power $\\$ $300 \times 10^6 = n \times 200 \times 10^6 \times 1.6 \times 10^{–19}$ $\\$ $\Rightarrow n = \dfrac{3}{2 \times 1.6} \times 10^{19} = \dfrac{3}{3.2} \times 10^{19}$ $\\$ $6 \times 10^{23}$ atoms are present in $238$ grams $\\$ $\dfrac{3}{3.2} \times 10^{19}$ atoms are present in $\dfrac{238 \times 3 \times 10^{19}}{6 \times 10^{23} \times 3.2} = 3.7 \times 10^{–4}\ g = 3.7\ mg.$

50.   A town has a population of 1 million. The average electric power needed per person is $300\ W$. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at $25$%. (a) Assuming $200\ MeV$ of thermal energy to come from each fission event on an average, find the number of events that should take place every day. (b) Assuming the fission to take place largely through $^{235}U$, at what rate will the amount of U decrease ? Express your answer in $kg/day$. (c) Assuming that uranium enriched to $3$% in $^{235}U$ will be used, how much uranium is needed per month $(30 days)$ ?

a) Energy radiated per fission $= 2 \times 10^8\ ev$ $\\$ Usable energy $= 2 \times 10^8 \times 25/100 = 5 \times 107 ev = 5 \times 1.6 \times 10^{–12}$ $\\$ Total energy needed $= 300 \times 10^8 = 3 \times 10^8\ J/s$ $\\$ No. of fission per second $= \dfrac{3 \times 10^8}{5 \times 1.6 \times 10^{-12}} = 0.375 \times 10^{20}$ $\\$ No. of fission per day $= 0.375 \times 10^{20} \times 3600 \times 24 = 3.24 \times 10^{24}$ fissions

b) From ‘a’ No. of atoms disintegrated per day $= 3.24 \times 10^{24}$ $\\$ We have, $6.023 \times 10^{23}$ atoms for $235\ g$ $\\$ for $3.24 \times 10^{24}\ atom = \dfrac{235}{6.023 \times 10^{24}} \times 3.24 \times 10^{24}\ g = 1264\ g/day = 1.264\ kg/day.$

51.   Calculate the $Q$-values of the following fusion, reactions: $\\$ (a) $^2H + ^2_{1}H \rightarrow\ _{1}H + ^1_{1}H$ $\\$ (b) $^2_{1}H + ^2H \rightarrow\ _2He + n$ $\\$ (c) $^2_{1}H + ^3_{1}H \rightarrow ^4_{2}He + n.$ $\\$ Atomic masses are $m(^2H) = 2\cdot014102\ u,\ m(^3H) = 3\cdot016049\ u,\ m(^3He) = 3\cdot016029\ u,\ m(^4He) = 4\cdot002603\ u.$

a) $^2_{1}H + ^2_{1}H \rightarrow ^3_{1}H + ^1_{1}H$ $\\$ Q value $= 2M(^2_{1}H) = [M(^2_{1}H) + M(^3_{1}H)]$ $\\$ $= [2 \times 2.014102 – (3.016049 + 1.007825)]u = 4.0275\ Mev = 4.05\ Mev.$

b) $^2_{1}H + ^2_{1}H \rightarrow ^3_{2}H + n$ $\\$ Q value $= 2[M(^2_{1}H) - M(^3_{2}He) + Mn]$ $\\$ $= [2 \times 2.014102 – (3.016049 + 1.008665)]u = 4.0275\ Mev = 3.25\ Mev.$

c) $^2_{1}H + ^3_{1}H \rightarrow ^4_{2}H + n$ $\\$ Q value $= 2[M(^2_{1}H) - M(^3_{2}He) - M(^4_{2}He) + Mn]$ $\\$ $= (2.014102 + 3.016049) – (4.002603 + 1.008665)]u = 17.58\ Mev = 17.57\ Mev$

52.   Consider the fusion in helium plasma. Find the temperature at which the average thermal energy $1\cdot5\ kT$ equals the Coulomb potential energy at $2\ fm.$

$PE = \dfrac{Kq_1q_2}{r} = \dfrac{9 \times 10^9 \times (2 \times 1.6 \times 10^{-19})^2}{r} \ \ \ \ \ \ \ \ \ \ \ \ \ \ …(1)$ $\\$ $1.5\ KT = 1.5 \times 1.38 \times 10^{–23} \times T \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ …(2)$

Equating $(1)$ and $(2)$ $1.5 \times 1.38 \times 10^{–23} \times T = \dfrac{9 \times 10^9 \times 10.24 \times 10^{-38}}{2 \times 10^{-15}}$

$\Rightarrow\ T = \dfrac{9 \times 10 \times 10.24 \times 10^{-38}}{2 \times 10^{-15} \times 1.5 \times 1.38 \times 10^{-23}} = 22.26087 \times 10^9\ K = 2.23 \times 10^{10}\ K.$

53.   Calculate the Q-value of the fusion reaction $$^4He + ^4He =\ ^8Be.$$ Is such a fusion energetically favourable ? Atomic mass of 8Be is 8 0053 u and that of $^4He$ is $4\cdot0026\ u.$

$^4H + ^4H \ \ \ \ \ \ \ \rightarrow ^8Be$ $\\$ $M(^2H) \ \ \ \ \ \ \ \rightarrow 4.0026\ u$ $\\$ $M(^8Be) \ \ \ \ \ \rightarrow 8.0053\ u$ $\\$ Q value $= [2 M(^2H) – M(^8Be)] = (2 \times 4.0026 – 8.0053)\ u$ $\\$ $= –0.0001\ u = –0.0931\ Mev = –93.1\ Kev.$

54.   Calculate the energy that can be obtained from $1$ kg of water through the fusion reaction $$^2H + ^2H \rightarrow ^3H + p.$$ Assume that $1\cdot5 \times 10^{-2}$% of natural water is heavy water $D_2O$ (by number of molecules) and all the deuterium is used for fusion.

In $18\ g$ of $N_0$ of molecule $= 6.023 \times 10^{23}$ $\\$ In $100\ g$ of $N_0$ of molecule $= \dfrac{6.023 \times 10^{26}}{18} = 3.346 \times 10^{25}$ $\\$ $\therefore$ % of Deuterium $= 3.346 \times 1026 \times 99.985$ $\\$ Energy of Deuterium $= 30.4486 \times 10^{25} = (4.028204 – 3.016044) \times 93$ $\\$ $= 942.32\ ev = 1507 \times 10^5\ J = 1507\ mJ$