# The Nucleus

## Concept Of Physics

### H C Verma

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 ?

##### Solution :

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

##### Solution :

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

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

##### Solution :

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 $^4He - 4\cdot0026\ u$.

##### Solution :

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

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

##### Solution :

$B = (Zm_p + Nm_n – M)C^2$ $\\$ $Z = 79 ;\ N = 118 ;\ mp = 1.007276u ;\ M = 196.96\ u ;\ mn = 1.008665u$ $\\$ $B = [(79 \times 1.007276 + 118 \times 1.008665)u – Mu]c^2$ $\\$ $= 198.597274 \times 931 – 196.96 \times 931 = 1524.302094$ $\\$ 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.

##### Solution :

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:$\\$ $$^{223}Ra \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{209}Pb \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{14}C$$ $\\$ $$223\cdot018\ u \ \ \ \ \ \ \ \ \ 208\cdot981\ u \ \ \ \ \ \ \ \ \ 14\cdot003\ u$$

##### Solution :

$^{223}R_a = 223.018\ u ;\ ^{209}Pb = 208.981\ u ;\ ^{14}C = 14.003\ u$. $\\$ $^{223}R_a \rightarrow ^{209}Pb + ^{14}C$ $\\$ $\Delta{m} = mass\ ^{223}R_a\ –\ mass\ (^{209}Pb + ^{14}C)$ $\\$ $\Rightarrow = 223.018 – (208.981 + 14.003) = 0.034.$ $\\$ 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 $proton-separation\ energy.$

##### Solution :

$E_{Z.N}. \rightarrow E_{Z–1},\ N + P_1 \Rightarrow E_{Z.N}. \rightarrow E_{Z–1},\ N + _1H^1$ [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.

##### Solution :

$E_{2}N = E_{Z,N–1} + ^{1}_{0}n$. $\\$ 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 $^{32}S = 31\cdot972\ u.$

##### Solution :

$P^{32} \rightarrow S^{32}\ +\ _0v^{-0}\ +\ _1\beta^0$ $\\$ 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.

##### Solution :

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) $^{25}_{13}AI\ \rightarrow \ ^{25}_{12}MS\ +$

##### Solution :

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

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.

##### Solution :

$^{64}Cu \rightarrow\ ^{64}Ni + e^- + v$ $\\$ 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.

##### Solution :

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

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

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.

##### Solution :

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

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 $^{11}B$.

##### Solution :

$"C \rightarrow "B + \beta^+ + v$ $\\$ 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: $$^{228}Th\ \rightarrow\ ^{224}Ra^ * + \alpha$$ $$^{224}Ra^*\ \rightarrow\ ^{224}Ra + \gamma\ (217\ keV).$$

##### Solution :

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 $12\cdot018613\ u.$

##### Solution :

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 ?

##### Solution :

$^{12}N \rightarrow ^{12}C* + e^+ + v$ $\\$ $^{12}C* \rightarrow ^{12}C + v(4.43\ Mev)$ $\\$ 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 $198\ g/mol.$

##### Solution :

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 $198\ g/mol.$

##### Solution :

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

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 $\\$

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 ?

##### Solution :

$t_{1}{2} = 8.0\ days ; A_0 = 20 \mu\ Cl$ $\\$ 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 ?

##### Solution :

$\lambda = 4.9 \times 10^{–18}\ s^{–1}$ $\\$ 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 ?

##### Solution :

$A = 200,\ A_0 = 500,\ t = 50\ min$ $\\$ $A = A_0 e^{–\lambda{t}}$ 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 ?

##### Solution :

$A_0 = 4 \times 10^5\ disintegration / sec$ $\\$ $A' = 1 \times 10^6\ dis/sec ;\ t = 20\ hours.$ $\\$ $A' = \dfrac{A_0}{2^{t / t_{1/ 2}}} \Rightarrow 2^{t / t_{1/ 2}} = \dfrac{A_0}{A'} \Rightarrow 2^{t / t_{1/ 2}} = 4$ $\\$ $\Rightarrow t / t_{1/ 2} = 2 \Rightarrow t_{1/2} = \dfrac{t}{2} = \dfrac{20\ hours}{2} = 10\ hours.$ $\\$ $A" = \dfrac{A_0}{2^{t / t _{1/ 2}}} \Rightarrow A" = \Rightarrow \dfrac{4 \times 10^6}{2^{100 /10}} = 0.00390625 \times 10^6 = 3.9 \times 10^3$ 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 $35\cdot5\ g/mol.$

##### Solution :

$t_{1/2} = 1602\ Y ;\ Ra = 226\ g/mole ;\ Cl = 35.5\ g/mole.$ $\\$ $1\ mole\ RaCl_2 = 226 + 71 = 297\ g$ $\\$ $297g = 1 mole of Ra.$ $\\$ $0.1\ g = \dfrac{1}{297} \times 0.1 mole of Ra = \dfrac{0.1 \times 6.023 \times 10^{23}}{297} = 0.02027 \times 10^{22}$ $\\$ $\lambda = \dfrac{0.693}{t_{1/2}} = 1.371 \times 10{–11}.$ $\\$ 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 $1\ Ci.$

##### Solution :

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

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 $\\$ $A_10 = \lambda N_10$

$t_{1/2} = 10$ 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, $A_9 = \lambda N_9$ $\\$ $\Rightarrow N_9 = {A_9}{\lambda} = \dfrac{0.536 \times 10 \times 3.7 \times 10^{10} \times 3600}{0.693}= 28.6176 \times 10^{10} \times 3600 = 103.023 \times 10^{13}.$

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 ?

##### Solution :

$t_{1/2} = 14.3\ days ;\ t = 30\ days = 1 month$ $\\$ 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 ?

##### Solution :

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-$11$ ?

##### Solution :

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] $\\$ $\Rightarrow In \dfrac{1}{9} = \dfrac{-0.693}{20.3}t \Rightarrow t = \dfrac{2.1972 \times 20.3}{0.693} = 64.36 = 64\ min.$

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 $10\ N_0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 90\ N_0$

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 $6\cdot15\ y.$

##### Solution :

$= 7.146 \times 10^{14}$ 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 $= (4 – 2.82) \times 10^{23} = 1.18 \times 10^{23}.$

$N = 4 \times 10^{23} ;\ t_{1/2} = 12.3$ 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

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.

##### Solution :

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.

##### Solution :

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

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

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 $U^{238} =\dfrac{6 \times 10^{23} \times 2 \times 10^{-3}}{238} = \dfrac{12}{238} \times 10^{20} = 0.05042 \times 10^{20}$

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 $5730\ y.$

##### Solution :

$A_0 = 15.3 ;\ A = 12.3 ;\ t_{1/2} = 5730\ year$ $\\$ $\lambda = \dfrac{0.6931}{T_{1/ 2}} \dfrac{0.6931}{5730} yr^{-1}$ $\\$ 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.

##### Solution :

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 $T_{\frac{1}{2}}$.

##### Solution :

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

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) $In(R_0/R_5) = In \Big(\dfrac{30 \times 10^9}{2 \times 10^9}\Big) = 2.7$

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.