NCERT Solutions Class 12 Physics Chapter 13

Q.1 (a) Two stable isotopes of lithiumLi and36Li have respective37abundance of 7.5 % and 92.5 %. These isotopes have masses6.01512 u and 7.01600 u respectively. Find the atomic weightof lithium.(b) Boron has two stable isotopes B and510B. Their respective511masses are 10.01294 u and 11.00931 u and the atomic mass ofboron is 10.811 u. Find the abundances ofB and510B.511

Ans.

(a) Atomic weight of lithium can be calculated as

\begin{array}{l}\text{The}\text{atomic}\text{weight}\text{is}\text{the}\text{weighted}\text{average}\text{of}\text{the}\text{isotopes}\text{.}\\ \text{atomic}\text{weight}=\frac{\text{6}.0\text{1512}\times \text{7}.\text{5}+\text{7}.0\text{16}00\times \text{92}.\text{5}}{\text{7}.\text{5}+\text{92}.\text{5}}\\ \text{atomic}\text{weight}=\frac{\text{45}.\text{1134}+\text{648}.\text{98}}{\text{1}00}\\ \text{atomic}\text{weight}=\text{6}.\text{941 u}\end{array}

(b) Let x% and (100-x)% be the abundances of ₅B¹⁰ and ₅B¹¹ respectively

$\begin{array}{l}\text{The atomic weight of Boron can be written as}\\ 10.811=\frac{\mathrm{}[\text{x\hspace{0.17em}}(\text{1}0.0\text{1294})+\text{(100\hspace{0.17em}-\hspace{0.17em}x)}(\text{11}.00\text{931})]}{\text{1}00}\\ 10.811=\frac{\mathrm{}[\text{1}0.0\text{1294\hspace{0.17em}x}+\text{1100.931-11}.00\text{931\hspace{0.17em}x}]}{\text{1}00}\\ 10.811=\frac{\mathrm{}[\text{1100.931-}0.99637\text{\hspace{0.17em}x}]}{\text{1}00}\\ 1081.1=\text{1100.931\hspace{0.17em}-\hspace{0.17em}}0.99637\text{\hspace{0.17em}x}\\ \mathrm{x}=19.91\mathrm{\%}\\ \text{and (100\hspace{0.17em}-\hspace{0.17em}x)}=\text{1}00–\text{19}.\text{91}\\ =\text{8}0.0\text{9}\mathrm{\%}\\ \text{Thus,\hspace{0.17em}19.9\hspace{0.17em}\%\hspace{0.17em}and\hspace{0.17em}80.09\hspace{0.17em}\%\hspace{0.17em}are\hspace{0.17em}the\hspace{0.17em}abudances\hspace{0.17em}of\hspace{0.17em}}\text{B}_{\text{10}}^{\text{5}}\text{\hspace{0.17em}and\hspace{0.17em}}\text{B}_{\text{11}}^{\text{5}}\text{.}\end{array}$

Q.2 The three stable isotopes of neon:Ne,1020Ne and1021Ne have1022respective abundance of 90.51 %, 0.27 % and 9.22 %. Theatomic masses of three isotopes are 19.99 u, 20.99 u and 21.99 u,respectively. Obtain the average atomic mass of neon.

Ans.

$\begin{array}{l}\text{As average atomic mass of neon can be obtained as}\\ \text{at.\hspace{0.17em}mass}=\frac{(\text{9}0.\text{51}\times \text{19}.\text{99})+(0.\text{27}\times \text{2}0.\text{99})+(\text{9}.\text{22}\times \text{21}.\text{99})}{\text{1}00}\mathrm{}\\ \text{at.\hspace{0.17em}mass}\mathrm{}=\frac{\text{18}0\text{9}.\text{29}+\text{5}.\text{67}+\text{2}0\text{2}.\text{75}}{\text{1}00\mathrm{}}\\ =\text{2}0.\text{18 u\hspace{0.17em}.}\end{array}$

Q.3

\begin{array}{l}\text{Obtain the binding energy}\left(\text{in MeV}\right)\text{of a nitrogen nucleus (}\text{N}_{\text{7}}^{\text{14}}\text{)}\text{.}\\ \text{given m (}\text{N}_{\text{7}}^{\text{14}}\text{)=14}\text{.00307 u}\text{.}\end{array}

Ans.

\begin{array}{l}\text{Mass defect of nucleus of Nitrogen will be}\\ \Delta \text{m}=\left[{\text{Zm}}_{\text{H}}+\left(\text{A}-\text{Z}\right){\text{m}}_{\text{n}}\right]–{\text{m}}_{\text{N}}\\ \text{where}\text{A}=\text{14},\text{Z}=\text{7 thus}\\ \Delta \text{m}=\left[\text{7}\left(\text{1}.00\text{783}\right)+\left(\text{14}-\text{7}\right)\text{1}.00\text{867}\right]–\text{14}.00\text{3}0\text{7}\\ \Delta \text{m}=\text{7}.0\text{5481}+\text{7}.0\text{6}0\text{69}-\text{14}.00\text{3}0\text{7}\\ =0.\text{11243 u}\\ \Delta \text{m}=0.\text{11243 u}\\ \text{Thus B}\text{.E of the nucleus will be}\\ B.E=0.\text{11243}\times \text{931}\\ =\text{1}0\text{4}.\text{7M}\text{eV}\text{.}\end{array}

Q.4

$\begin{array}{l}\text{Obtain the binding energy of the nuclei\hspace{0.33em}}\text{Fe\hspace{0.33em}and\hspace{0.33em}}_{\text{26}}^{\text{56}}\text{Bi\hspace{0.33em}in units of}_{\text{83}}^{\text{209}}\\ \text{MeV from the following data:\hspace{0.33em}}\\ \text{m}\left(\text{Fe}_{\text{26}}^{\text{56}}\right)\text{= 55.934939 u\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} m}\left(\text{Bi}_{\text{83}}^{\text{209}}\right)\text{= 208.980388 u.}\end{array}$

Ans.

$\begin{array}{l}\text{For the nucleus of}\text{Fe, mass defect can be expressed as}_{\text{26}}^{\text{56}}\\ \text{Δm=}\left[{\text{Zm}}_{\text{H}}\text{+}\left(\text{A-Z}\right){\text{m}}_{\text{n}}\right]\text{–m}\left(\text{Fe}_{\text{26}}^{\text{56}}\right)\\ \text{Δm=}\left[\text{26}\left(\text{1.007825}\right)\text{+}\left(\text{56-26}\right)\text{\hspace{0.17em} 1.008665}\right]\text{–55.934939}\\ \text{Δm=0.528461 u\hspace{0.17em}.}\\ \text{Thus, binding energy}\\ \text{B.E=Δm×931 MeV}\\ \text{=492\hspace{0.17em}MeV}\\ \text{Binding energy per nucleon of Fe nucleus will be}\\ \text{=}\frac{\text{B.E}}{\text{A}}\\ \text{=}\frac{\text{492}}{\text{56}}\\ \text{=8.79 MeV.}\end{array}$

\begin{array}{l}\text{Similarly, for the nucleus of}\text{B}_{\text{83}}^{\text{209}}\text{i,}\\ \text{mass defect can be expressed as}\\ \text{Δm=}\left[{\text{Zm}}_{\text{H}}\text{+}\left(\text{A-Z}\right){\text{m}}_{\text{n}}\right]\text{-m}\left(\text{B}_{\text{83}}^{\text{209}}\text{i}\right)\\ \text{}\text{}\text{}\text{}\text{=}\left[\text{83}\left(\text{1}\text{.007825}\right)\text{+}\left(\text{209–83}\right)\text{1}\text{.008665}\right]\text{–208}\text{.980388}\\ \text{=1}\text{.760877 u}\text{.}\\ \text{Thus binding energy}\\ \text{B}\text{.E=Δm×931MeV}\\ \text{=1}\text{.760877×931}\\ \text{=1639}\text{.37}\text{MeV}\\ \text{Hence, binding energy per nucleon}\\ \text{of Bi nucleus will be=}\frac{\text{B}\text{.E}}{\text{A}}\\ \text{=}\frac{\text{1639}\text{.37}}{\text{209}}\\ \text{=7}\text{.84}\text{MeV}\text{.}\\ \text{F}_{\text{26}}^{\text{56}}\text{e}\text{has more Binding energy per nucleon compared to}\text{B}_{\text{83}}^{\text{209}}\text{i}\text{.}\end{array}

Q.5

\begin{array}{l}\text{A given coin has a mass of 3}\text{.0 g}\text{. Calculate the nuclear energy}\\ \text{that would be required to separate all the neutrons and protons}\\ \text{from each other}\text{. For simplicity assume that the coin is entirely}\\ \text{made of}\text{C}_{\text{29}}^{\text{63}}\text{u}\text{atoms}\left(\text{of mass 62}\text{.92960 u}\right)\text{.}\end{array}

Ans.

\begin{array}{l}\text{As mass defect is expressed as}\\ \Delta \text{m}=\left[{\text{Zm}}_{\text{p}}+\left(\text{A}-\text{Z}\right){\text{m}}_{\text{n}}\right]–\text{M}\\ \text{Here,}\\ \text{A}=\text{63},\text{Z}=\text{29},{\text{m}}_{\text{p}}=\text{1}.00\text{783 u},{\text{m}}_{\text{n}}=\text{1}.00\text{867}uand\\ \text{M}=\text{62}.\text{9296}0\text{u}\\ \text{Hence, mass defect}\\ \Delta \text{m}=\left[\text{29}\left(\text{1}.00\text{783}\right)+\text{34}\left(\text{1}.00\text{867}\right)\right]–\text{62}.\text{9296}0\\ \Delta \text{m}=0.\text{59225 u}\text{.}\\ \text{Thus, binding energy of Cu nucleus}\text{is}\\ B.E=\Delta \text{m}\times \text{931 MeV}\\ =0.\text{59225}\times \text{931}\\ B.E=\text{551}.\text{38 MeV}\text{.}\\ \text{As}n\text{umber of atoms in 63g of copper}=6.023\times {10}^{23}\\ N\text{umber of atoms in 1g of copper}=\frac{6.023\times {10}^{23}}{63}\\ N\text{umber of atoms in 3g of copper}=\frac{6.023\times {10}^{23}\times 3}{63}\\ =2.868\times {10}^{22}\\ \text{If binding energy of one atom}=\text{551}.\text{38 MeV}\\ \text{Binding energy of 2}.\text{868}\times \text{1}{0}^{\text{22}}\text{atoms}=\text{551}.\text{38}\times \text{2}.\text{868}\times \text{1}{0}^{\text{22}}MeV\\ =\text{1}.\text{58}\times \text{1}{0}^{\text{25}}\text{MeV}\\ \text{Thus},\text{This much energy is required to seprate all the neutrons}\\ \text{and protons from each other}\text{.}\end{array}

Q.6

$\begin{array}{l}\text{Write nuclear reaction equations for:}\\ \left(\text{i}\right)\text{\hspace{0.17em}\hspace{0.17em}α\hspace{0.17em}decay of \hspace{0.17em}}\text{Ra}_{\text{88}}^{\text{226}}\\ \left(\text{ii}\right)\text{\hspace{0.17em}α\hspace{0.17em}decay of \hspace{0.17em}}\text{Pu}_{\text{94}}^{\text{242}}\\ \left(\text{iii}\right){\text{\hspace{0.17em} β}}^{\text{–}}\text{decay of \hspace{0.17em}}\text{P}_{\text{15}}^{\text{32}}\\ \left(\text{iv}\right){\text{\hspace{0.17em} β}}^{\text{–}}\text{decay of}\text{Bi}_{\text{83}}^{\text{210}}\\ \left(\text{v}\right){\text{\hspace{0.17em} β}}^{\text{+}}\text{decay of}\text{C}_{\text{6}}^{\text{11}}\\ \left(\text{vi}\right){\text{\hspace{0.17em}β}}^{\text{+}}\text{decay of}\text{Tc}_{\text{43}}^{\text{97}}\\ \left(\text{vii}\right)\text{\hspace{0.17em} Electron capture of}\text{Xe}_{\text{54}}^{\text{120}}\end{array}$

Ans.

$\begin{array}{l}\text{In alpha decay, mass number decreases by 4 and charge number decreases by 2.}\\ \text{In beta minus decay, mass number remains unaffected and charge number increases by}\\ \text{one and there is emission of anti\hspace{0.17em}-neutrino from the nucleus.}\\ \left(\text{i}\right)\text{Ra}_{88}^{226}\stackrel{}{\to }\mathrm{}\text{Ra}_{86}^{222}+\text{He}_{\text{2}}^{\text{4}}\mathrm{}\\ \mathrm{}\left(\text{ii}\right)\text{Pu}_{94}^{242}\mathrm{}\stackrel{}{\to }\text{U}_{92}^{238}+\text{He}_2^4\mathrm{}\\ \mathrm{}\left(\text{iii}\right)\text{P}_{15}^{32}\stackrel{}{\to }\text{S}_{16}^{32}\mathrm{}+{\mathrm{e}}^{–}\mathrm{}+\overline{\mathrm{v}}\mathrm{}\text{(antineutrino)}\\ \left(\text{iv}\right)\text{Bi}_{83}^{210}\stackrel{}{\to }\text{Po}_{84}^{210}\mathrm{}+{\mathrm{e}}^{–}+\mathrm{}\overline{\mathrm{v}}\mathrm{}\text{(antineutrino)}\mathrm{}\\ \mathrm{}\left(\text{v}\right)\text{C}_6^{11}\mathrm{}\stackrel{}{\to }\text{B}_5^{11}+{\mathrm{e}}^{+}\text{(positron)}+\mathrm{v}\left(\text{neutrino}\right)\\ \left(\text{vi}\right)\text{Tc}_{43}^{97}\stackrel{}{\to }\text{Tc}_{43}^{97}+{\mathrm{e}}^{+}\left(\text{positron}\right)+\mathrm{v}\left(\text{neutrino}\right)\\ \left(\text{vii}\right)\text{Xe}_{54}^{120}\mathrm{}+{\mathrm{e}}^{+}\stackrel{}{\to }\text{I}_{53}^{120}+\mathrm{v}\text{(neutrino)}\end{array}$

Q.7 A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to
(a) 3.125 % of its original activity (b) 1% of original value ?

Ans.

(a) The rate of disintegration or count rate of a sample of a radioactive material is called activity (A) of the sample, which is directly proportional to the number of atoms left undecayed in the sample.

$\begin{array}{l}\text{Thus,}\\ \frac{\text{A}}{{\text{A}}_0}=\frac{\text{N}}{{\text{N}}_0}=\mathrm{}{\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{\mathrm{T}}}.\dots \dots \dots \dots \dots \dots \dots .\left(1\right)\\ \mathrm{As},{\text{\hspace{0.17em}A}}_0={\mathrm{N}}_0\mathrm{\lambda }\\ =\text{initial\hspace{0.17em}activity\hspace{0.17em}or\hspace{0.17em}decay\hspace{0.17em}rate\hspace{0.17em}at\hspace{0.17em}}\mathrm{t}=0\mathrm{and}\\ \text{A}=\mathrm{N}\mathrm{\lambda }\\ =\text{\hspace{0.17em}activity\hspace{0.17em}or\hspace{0.17em}decay\hspace{0.17em}rate\hspace{0.17em}at\hspace{0.17em}time\hspace{0.17em}t.}\\ {\text{Here,\hspace{0.17em}N}}_{\text{0}}\text{\hspace{0.17em}is\hspace{0.17em}original\hspace{0.17em}amount\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}radioactive\hspace{0.17em}isotope\hspace{0.17em}and}\\ \text{N\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}amount\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}radioactive\hspace{0.17em}isotope\hspace{0.17em}after\hspace{0.17em}decay.}\\ \text{According\hspace{0.17em}to\hspace{0.17em}question,}\\ \frac{\text{N}}{{\text{N}}_0}=3.125\mathrm{\%}\\ =\frac{3.125}{100}\\ =\frac{1}{32}\\ \frac{\text{N}}{{\text{N}}_0}=\frac{1}{32},\\ {\text{from\hspace{0.17em}eq}}^{\text{n}}\text{(1),\hspace{0.17em}we\hspace{0.17em}get}\\ \frac{\text{A}}{{\text{A}}_0}=\frac{1}{32}\\ =\mathrm{}{\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{\mathrm{T}}}\\ \frac{\text{A}}{{\text{A}}_0}={\left(\frac{1}{2}\right)}^5\\ =\mathrm{}{\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{\mathrm{T}}}\\ \frac{\mathrm{t}}{\mathrm{T}}=\mathrm{n};\frac{\mathrm{t}}{\mathrm{T}}=5\\ \mathrm{t}=5\mathrm{T}\\ \text{Hence,\hspace{0.17em}the\hspace{0.17em}isotope\hspace{0.17em}will\hspace{0.17em}take\hspace{0.17em}about\hspace{0.17em}5\hspace{0.17em}years\hspace{0.17em}to\hspace{0.17em}reduce\hspace{0.17em}to\hspace{0.17em}3.125\%\hspace{0.17em}of\hspace{0.17em}its\hspace{0.17em}orginal\hspace{0.17em}value.}\\ \left(\text{b}\right){\text{\hspace{0.17em}According\hspace{0.17em}to\hspace{0.17em}question,\hspace{0.17em}only\hspace{0.17em}1\%\hspace{0.17em}of\hspace{0.17em}N}}_{\text{0}}\text{\hspace{0.17em}remains\hspace{0.17em}after\hspace{0.17em}decay.}\\ \mathrm{Hence},\text{\hspace{0.17em}}\frac{\text{N}}{{\text{N}}_0}=1\mathrm{\%}\\ =\frac{1}{100}\\ \frac{\text{N}}{{\text{N}}_0}=\frac{1}{100},\\ \mathrm{But}\frac{\text{N}}{{\text{N}}_0}={\mathrm{e}}^{-\mathrm{\lambda t}}\\ =\frac{1}{100}\\ {\mathrm{e}}^{+\mathrm{\lambda t}}=100\\ \mathrm{\lambda t}={\log }_{\mathrm{e}}100\\ \mathrm{\lambda t}=2.3026\log 100\\ \mathrm{\lambda t}=4.6052\\ \mathrm{t}=\frac{4.6052}{\mathrm{\lambda }}\\ \mathrm{Since},\text{\hspace{0.17em}}\mathrm{\lambda }=\frac{0.693}{\mathrm{T}}\\ \mathrm{t}=\frac{4.6052\times \mathrm{T}}{0.693}\\ \mathrm{t}=6.65\mathrm{T}\\ \text{Hence,\hspace{0.17em}the\hspace{0.17em}isotope\hspace{0.17em}will\hspace{0.17em}take\hspace{0.17em}about\hspace{0.17em}6.65\hspace{0.17em}years\hspace{0.17em}to\hspace{0.17em}reduce\hspace{0.17em}to\hspace{0.17em}1\%\hspace{0.17em}of\hspace{0.17em}its\hspace{0.17em}orginal\hspace{0.17em}value.}\end{array}$

Q.8

$\begin{array}{l}\text{This activity arises from the small proportion of radioactive}\text{C}_{\text{6}}^{\text{14}}\\ \text{present with the stable carbon isotope}\text{C.\hspace{0.33em}When the organism is}_{\text{6}}^{\text{12}}\\ \text{dead, its interaction with the atmosphere}\\ \left(\text{which maintains the above equilibrium activity}\right)\text{ceases and its}\\ \text{activity begins to drop. From the known half-life}\left(\text{5730 years}\right)\\ \text{of\hspace{0.33em}}\text{C, and the measured activity, the age of the specimen can be}_{\text{6}}^{\text{14}}\\ {\text{approximately estimated.\hspace{0.33em}This is the principle of\hspace{0.33em}}}_{\text{6}}{\text{C}}^{\text{14}}\text{dating used in}\\ \text{archaeology. Suppose a specimen from Mohenjodaro gives an activity}\\ \text{of 9 decays per minute per gram of carbon. Estimate the approximate}\\ \text{age of the Indus-Valley civilization.}\end{array}$

Ans.

$\begin{array}{l}{\text{Here normal activity, R}}_0=\text{15 decays/min and activity of a specimen from Mohenjodaro,}\\ \text{R}=\text{9 decays}/\text{min}.\\ \mathrm{As}\mathrm{activity}\mathrm{is}\mathrm{proportional}\mathrm{to}\mathrm{the}\text{\hspace{0.17em}}\mathrm{number}\mathrm{of}\mathrm{radioactive}\mathrm{atoms},\\ \frac{\mathrm{N}}{{\mathrm{N}}_0}=\frac{\mathrm{R}}{{\mathrm{R}}_0}=\frac{9}{15}.\dots \dots \dots \dots \dots .\left(1\right)\\ \text{From\hspace{0.17em}}\mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}},\\ \frac{\mathrm{N}}{{\mathrm{N}}_0}={\mathrm{e}}^{-\mathrm{\lambda t}}\\ \text{From\hspace{0.17em}}{\mathrm{eq}}^{\mathrm{n}}\left(1\right)\\ \frac{\mathrm{N}}{{\mathrm{N}}_0}={\mathrm{e}}^{-\mathrm{\lambda t}}\\ =\frac{9}{15}\\ {\mathrm{e}}^{-\mathrm{\lambda t}}=\frac{9}{15}\\ \mathrm{e}{}^{\mathrm{\lambda }\mathrm{t}}=\frac{15}{9}\\ \mathrm{\lambda }\mathrm{t}=2.3026\log (1.666)\\ \mathrm{\lambda }\mathrm{t}=0.5109\\ \mathrm{t}=\frac{0.5109}{\mathrm{\lambda }}\\ \text{Since,\hspace{0.17em}}\mathrm{\lambda }=\frac{0.693}{\mathrm{T}}\\ \mathrm{T}=5730\mathrm{year}\left(\text{Given}\right)\\ \mathrm{t}=\frac{0.5109\times \mathrm{T}}{0.693}\\ \mathrm{t}=\frac{0.5109\times 5730}{0.693}\\ =4224.32\mathrm{years}\\ \text{Thus, the approximate age of the Indus\hspace{0.17em}Valley}\\ \text{Civilization is 4225 years\hspace{0.17em}.}\end{array}$

Q.9

$\begin{array}{l}\text{Obtain the amount of}\text{Co\hspace{0.33em}necessary to provide a radioactive}_{\text{27}}^{\text{60}}\\ \text{source of 8.0 mCi strength. The half-life of}\text{Co\hspace{0.33em}is 5.3 years.}_{\text{27}}^{\text{60}}\end{array}$

Ans.

$\begin{array}{l}\text{As 1 Curie}=\text{3}.\text{7 x 1}{0}^{\text{1}0}\text{disintegrations per second}\\ \text{Hence}\\ \text{8}.0\text{mCi}\mathrm{}=\text{8}\times \text{1}{0}^{-\text{3}}\times \text{3}.\text{7}\times \text{1}{0}^{\text{1}0}\\ =\text{\hspace{0.17em}2}.\text{96}\times \text{1}{0}^{\text{8}}\mathrm{}{\text{disintegrations s}}^{\text{-1}}\\ \text{Also half life,}\\ {\text{T}}_{\text{1}/\text{2}}=\text{5}.\text{3 years}\\ =\text{5}.\text{3}\mathrm{}\times \text{365}\times \text{24}\times \text{6}0\times \text{6}0\text{s}\\ =\text{1}.\text{67}\times \text{1}{0}^{\text{8}}\text{s}\\ \text{As per radioactive decay law,}\mathrm{}\\ -\frac{\text{dN}}{\text{dt}}=\mathrm{\lambda }\text{N}\mathrm{}\mathrm{o}\text{r\hspace{0.17em}N}=\frac{\mathrm{}\text{1}}{\mathrm{\lambda }\mathrm{}}\mathrm{}(-\frac{\text{dN}}{\text{dt}})\\ \mathrm{N}=\left(\frac{{\text{T}}_{\frac{1}{2}}}{0.\text{693}}\right)\mathrm{}(-\frac{\text{dN}}{\text{dt}})\mathrm{},\text{\hspace{0.17em}\hspace{0.17em}as}\mathrm{}[{\text{T}}_{\frac{1}{2}}=\frac{0.\text{693}}{\mathrm{\lambda }}]\mathrm{}\\ \mathrm{}\text{Thus, N}=\left(\frac{\text{1}.\text{67}\times \text{1}{0}^{\text{8}}}{0.\text{693}}\right)\mathrm{}(\text{2}.\text{96}\times \text{1}{0}^{\text{8}})\mathrm{}\\ =\text{7}.\text{13}\times \text{1}{0}^{\text{16}}\\ \text{As\hspace{0.17em}}\mathrm{}\text{6}.0\text{23}\times \text{1}{0}^{\text{23}}\text{\hspace{0.17em}atoms of cobalt have mass}=\text{6}0\text{g}\\ \text{Thus\hspace{0.17em}7}.\text{13}\times \text{1}{0}^{\text{16}}\mathrm{}\text{atoms of cobalt have mass}=\frac{\text{6}0\times \text{7}.\text{13}\times \text{1}{0}^{\text{16}}}{\text{6}.0\text{23}\times \text{1}{0}^{\text{23}}}\mathrm{}\\ =\text{7}.\text{1}\times \text{1}{0}^{-\text{6}}\text{\hspace{0.17em}g}\end{array}$

Q.10

$\begin{array}{l}\text{The half life of}\text{Sr\hspace{0.33em}is 28 years. What is the disintegration rate}_{\text{38}}^{\text{90}}\\ \text{of 15 mg of this isotope?}\end{array}$

Ans.

$\begin{array}{l}\text{Given}:\\ \text{T}=\text{28 years}=\text{28}\times \text{365}\times \text{24}\times \text{36}00=\text{8}.\text{83}0\times \text{1}{0}^{-\text{8}}\\ \mathrm{As}\mathrm{Number}\mathrm{of}\text{\hspace{0.33em}atoms present in 90\hspace{0.33em}g of}\mathrm{Sr}_{38}^{90}=\text{6}.0\text{23}\times \text{1}{0}^{\text{23}}\\ \mathrm{Number}\mathrm{of}\text{\hspace{0.33em}atoms present in 1g of}\mathrm{Sr}_{38}^{90}=\frac{\text{6}.0\text{23}\times \text{1}{0}^{\text{23}}}{90}\\ \therefore \mathrm{number}\mathrm{of}\text{\hspace{0.33em}atoms present in 15mg of}\mathrm{Sr}_{38}^{90}=\frac{\text{6}.0\text{23}\times \text{1}{0}^{\text{23}}\times \text{15}}{90\times 1000}\\ =1.0038\times {10}^{20}\\ \text{As disintegration constant\hspace{0.33em}}\mathrm{\lambda },\text{\hspace{0.33em}is expressed as}\\ \mathrm{\lambda }=\frac{0.\text{693}}{{\text{T}}_{\frac{1}{2}}}\\ =\left[\frac{0.\text{693}}{\left(8.830\times {10}^{-8}\right)}\right]{\text{s}}^{-\text{1}}\\ =0.0784\times {10}^8{\text{s}}^{-\text{1}}\\ \text{Rate of disintegration}\\ \left(\frac{\text{dN}}{\text{dt}}\right)=\mathrm{\lambda }\text{N}\\ \left(\frac{\text{dN}}{\text{dt}}\right)=0.0784\times {10}^8\times 1.0038\times {10}^{20}\\ \left(\frac{\text{dN}}{\text{dt}}\right)=\text{7}.\text{87 x 1}{0}^{\text{1}0}\text{Bq}.\text{Or 2}.\text{13 Ci\hspace{0.17em}.}\end{array}$

Q.11

\begin{array}{l}\text{Obtain approximately the ratio of the nuclear radii of}\\ \text{the gold isotope}\text{A}_{\text{79}}^{\text{197}}\text{u}\text{and silver isotope}\text{A}_{\text{47}}^{\text{107}}\text{g}\text{.}\text{Obtain}\\ \text{approximately the ratio of the nuclear radii of the gold}\\ \text{isotope}\text{A}_{\text{79}}^{\text{197}}\text{u}\text{and silver isotope}\text{A}_{\text{47}}^{\text{107}}\text{g}\text{.}\end{array}

Ans.

\begin{array}{l}\text{As nuclear radius}\text{R}\propto {\text{A}}^{\frac{1}{3}}\text{where A is atomic mass of the nuclei}\text{.}\\ \frac{{\text{R}}_{\text{Au}}}{{\text{R}}_{\text{Ag}}}={\left(\frac{{\text{A}}_{\text{Au}}}{{\text{A}}_{\text{Ag}}}\right)}^{\frac{1}{3}}\\ ={\left(\frac{\text{197}}{\text{1}0\text{7}}\right)}^{\frac{1}{3}}\\ ={\left(\text{1}.\text{84}\right)}^{\frac{1}{3}}\\ =\text{1}.\text{23}\end{array}

Q.12

$\begin{array}{l}\text{Find the Q-value and the kinetic energy of the emitted\hspace{0.33em}particle}\\ \text{in the decay of}\left(\text{a}\right)\mathrm{Ra}_{88}^{226}\mathrm{and}\left(\text{b}\right)\mathrm{Rn}_{86}^{220}\\ \text{Given,}\\ \text{m}\left(\mathrm{Ra}_{88}^{226}\right)\text{= 226.02540 u;\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}m}\left(\mathrm{Rn}_{86}^{222}\right)\text{= 222.01750 u}\\ \mathrm{m}\left(\mathrm{Rn}_{86}^{220}\right)\text{= 220.01137 u;\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}m}\left(\mathrm{Po}_{84}^{216}\right)\text{= 216.00189 u.}\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Ra}_{\text{88}}^{\text{226}}\to \text{Rn+}_{\text{86}}^{\text{222}}\text{He}_{\text{2}}^{\text{4}}\\ \text{Thus, the value of binding energy or Q-Value}\\ \text{=m}\left(\text{Ra}_{\text{88}}^{\text{226}}\right)\text{–}\left[\text{m}\left(\text{Rn}_{\text{86}}^{\text{222}}\right)\text{+m}\left(\text{He}_{\text{2}}^{\text{4}}\right)\right]\\ \text{=226.02540–226.0201}\\ \text{=0.0053 u}\\ \text{Thus}\\ \text{Q=0.0053×931 MeV}\left(\text{as 1 u = 931 MeV}\right)\\ \text{Q=4.93 MeV}\\ {\text{The K.E of an\hspace{0.33em}α\hspace{0.33em}particle\hspace{0.33em}E}}_{\text{α}}\text{\hspace{0.33em}during decay of Ra will be}\\ {\text{E}}_{\text{α}}\text{=}\left(\frac{\text{Mass\hspace{0.33em}number\hspace{0.33em}after\hspace{0.33em}decay}}{\text{Mass\hspace{0.33em}number\hspace{0.33em}before\hspace{0.33em}decay}}\right)\text{×Q}\\ {\text{E}}_{\text{α}}\text{=}\left(\frac{\text{222}}{\text{226}}\right)\text{×4.93}\\ \text{=4.84 MeV\hspace{0.17em}.}\\ \left(\mathrm{b}\right)\text{Rn}_{\text{86}}^{\text{220}}------\to \text{Po}_{\text{82}}^{\text{216}}+\text{He}_2^4\\ \text{Q}-\text{Value}\\ =\text{m}\left(\text{Rn}_{\text{86}}^{\text{220}}\right)–\left[\text{m}\left(\text{Po}_{\text{82}}^{\text{216}}\right)+\text{m}\left(\text{He}_2^4\right)\right]\\ =\text{22}0.0\text{11373}–\left[\text{216}.00\text{189}+\text{4}.00\text{26}0\right]\\ =\text{22}0.0\text{11373}–\text{22}0.00\text{449}\\ =0.00\text{6883 u}\\ =0.00\text{6883}\times \text{931 MeV}\\ =\text{6}.\text{41 MeV\hspace{0.17em}.}\\ \text{K}.\text{E}.\text{of}\mathrm{\alpha }-\text{particle emitted during decay of Rn will be}\\ {\text{E}}_{\mathrm{\alpha }}=\left(\frac{\mathrm{Mass}\mathrm{number}\mathrm{after}\mathrm{decay}}{\mathrm{Mass}\mathrm{number}\mathrm{before}\mathrm{decay}}\right)\times \text{Q}\\ {\text{E}}_{\mathrm{\alpha }}=\left(\frac{\text{216}}{\text{220}}\right)\times \text{4}.\text{93}=\text{4}.\text{84 MeV}\end{array}$

Q.13

\begin{array}{l}\text{The radionuclide}\text{C}_{\text{6}}^{\text{11}}\text{decays according to}\text{C}_{\text{6}}^{\text{11}}\text{®}\text{B}_{\text{5}}^{\text{11}}{\text{+ e}}^{\text{+}}\text{+ v:}\\ \text{half life=20}\text{.3 min}\text{.}\text{The maximum energy of the emitted positron is 0}\text{.960 MeV}\text{.}\\ \text{Given the mass values}\text{.}\\ \text{m}\left(\text{C}_{\text{6}}^{\text{11}}\right)\text{= 11}\text{.011434 u; m}\left(\text{B}_{\text{5}}^{\text{11}}\right)\text{= 11}\text{.009305 u}\\ \text{Calculate Q and compare it with the maximum energy of the positron emitted}\text{.}\end{array}

Ans.

\begin{array}{l}\text{As given the equation of the decay process of}\text{C}_6{}^{11}\text{is}\\ \text{C}_6{}^{\text{11}}----\to {}_5{\text{B}}^{11}+{\text{e}}^{+}+v+\text{Q}\\ \text{Thus energy}\\ \text{Q}=\left[{\text{m}}_{\text{N}}\left(\text{C}_6{}^{\text{11}}\right)–{\text{m}}_{\text{N}}\left({}_5{\text{B}}^{11}\right)–{\text{m}}_{\text{e}}\right]\times \text{931}\text{MeV}\\ \text{Where the masses used are of nuclei and not of atoms}.\text{In order}\\ \text{to express the Q value in terms of the atomic masses},{\text{6 m}}_{\text{e}}\text{mass}\\ \text{has to be subtracted from atomic mass of carbon}{\text{and 5 m}}_{\text{e}}\text{mass}\\ \text{has to be subtracted from atomic mass of boron}\\ \left(\text{because their atoms contain outside electrons}\right).\text{Therefore},\\ \text{in terms of the atomic mass m},\text{the expression for energy will be}\\ \text{Q=}\left[\text{m}\left(\text{C}_{\text{6}}{}^{\text{11}}\right){\text{–6m}}_{\text{e}}\text{–m}\left({}_{\text{5}}{\text{B}}^{\text{11}}\right){\text{+5m}}_{\text{e}}{\text{–m}}_{\text{e}}\right]\times \text{931 MeV}\\ \text{=}\left[\text{m}\left(\text{C}_{\text{6}}{}^{\text{11}}\right)\text{–m}\left({}_{\text{5}}{\text{B}}^{\text{11}}\right){\text{-2m}}_{\text{e}}\right]\text{×931MeV}\\ \text{=}\left[\text{11}\text{.011434}\text{–}\text{11}\text{.009305}\text{–}\text{2}\times \text{0}\text{.000548}\right]\times \text{931}\\ \text{=0}\text{.001033}\times \text{931}\\ \text{=0}\text{.961 MeV}\text{.}\\ \text{This energy is comparable to actual energy released in the decay}\\ \text{process}.{\text{The daughter nucleus}}_{\text{5}}{\text{B}}^{\text{11}}\text{is very heavy as compared to}\\ {\text{e}}^{+}\text{and neutrino so the daughter nucleus carries minimum}\text{energy}.\\ \text{If the kinetic energy carried by the neutrino is minimum},\text{then}\\ \text{the positron carries maximum energy which is practically same as energy Q}.\end{array}

Q.14

$\begin{array}{l}\text{The nucleus}\text{Ne decays by- emission. Write down the-decay}_{\text{10}}^{\text{23}}\\ \text{equation and determine the maximum kinetic energy of the}\\ \text{electrons emitted.\hspace{0.33em}Given that:}\\ \text{m}\left(\text{Ne}_{\text{10}}^{\text{23}}\right)\text{= 22.994466 u}\\ \text{m}\left(\text{B}_{\text{5}}^{\text{11}}\right)\text{= 22.089770 u}\end{array}$

Ans.

$\begin{array}{l}\text{The decay process of\hspace{0.33em}}\text{Ne\hspace{0.33em}}_{10}^{23}\text{is}\\ \text{Ne}_{10}^{23}-----\to \text{Na}_{11}^{23}+{\text{e}}^{-}+\text{v}+\text{Q}\\ \text{If whole of energy is carried by}\mathrm{\beta }\text{particle}.\text{Then}\\ \text{Q}=\left[{\text{m}}_{\text{N}}\left(\text{Ne}_{10}^{23}\right)–{\text{m}}_{\text{N}}\left(\text{Na}_{11}^{23}\right)–{\text{m}}_{\text{e}}\right]\times \text{931}.\text{5 MeV}\\ \text{Where the masses used as}{\text{\hspace{0.33em}m}}_{\text{N}}\left(\text{Ne}_{10}^{23}\right)\text{\hspace{0.33em}and\hspace{0.33em}}{\text{m}}_{\text{N}}\left(\text{Na}_{11}^{23}\right)\text{\hspace{0.33em}are\hspace{0.33em}}\text{Ne}_{10}^{23}\\ \text{and}\text{Na\hspace{0.33em}respectively}_{11}^{23}.\\ \text{If\hspace{0.33em}m}\left(\text{Ne}_{10}^{23}\right)\text{and}\text{\hspace{0.33em}m}\left(\text{Na}_{11}^{23}\right)\text{\hspace{0.33em}are atomic\hspace{0.33em}masses of\hspace{0.33em}}\text{Ne}_{10}^{23}\text{\hspace{0.33em}and}\text{Na}_{11}^{23}\\ \text{respectively,\hspace{0.33em}}\mathrm{t}\text{hen}\\ {\text{m}}_{\text{N}}\left(\text{Ne}_{10}^{23}\right)=\text{m}\left(\text{Ne}_{10}^{23}\right)–\text{1}0{\text{m}}_{\text{e}}\text{\hspace{0.33em}as atom of Ne contains 1}0\text{electrons}.\\ {\text{Similarly\hspace{0.33em}m}}_{\text{N}}\left(\text{Na}_{11}^{23}\right)=\text{m}\left(\text{Na}_{11}^{23}\right)–{\text{11m}}_{\text{e}}\text{\hspace{0.33em}as atom of Na contains}\\ \text{11 electrons.}\\ \text{Thus,}\\ \text{Q}=\left[\text{m}\left(\text{Ne}_{10}^{23}\right)–\text{1}0{\text{m}}_{\text{e}}–\text{m}\left(\text{Ne}_{10}^{23}\right)+{\text{11m}}_{\text{e}}–{\text{m}}_{\text{e}}\right]\times \text{931}\\ \text{Q}=\left[\text{m}\left(\text{Ne}_{10}^{23}\right)–\text{m}\left(\text{Na}_{11}^{23}\right)\right]\text{x 931\hspace{0.17em}MeV}\\ \text{Q}=\left[\text{22}.\text{994466\hspace{0.33em}}–\text{\hspace{0.33em}22}.\text{98977}0\right]\times \text{931\hspace{0.17em}MeV}\\ \text{Q}=0.00\text{4689}\times \text{931}\\ =\text{4}.\text{37 MeV\hspace{0.17em}}.\end{array}$

Q.15

$\begin{array}{l}\text{The Q value of a nuclear reaction A + b = C + d is defined by Q =}\left[{\text{m}}_{\text{A}}{\text{+ m}}_{\text{b}}{\text{– m}}_{\text{C}}{\text{– m}}_{\text{d}}\right]{\text{c}}^{\text{2}}\\ \text{where the masses refer to the respective nuclei.}\\ \text{Determine from the given data the Q value of the following reactions and state whether the}\\ \text{following reactions are exothermic or endothermic.}\\ \text{(i) \hspace{0.17em}}\text{H +}_{\text{1}}^{\text{1}}\text{H}_{\text{1}}^{\text{3}}\stackrel{}{\to }\text{H +}_{\text{1}}^{\text{2}}\text{H}_{\text{1}}^{\text{2}}\\ \text{(ii)\hspace{0.17em}}\text{C +}_{\text{6}}^{\text{12}}\text{C}_{\text{6}}^{\text{12}}\stackrel{}{\to }\text{Ne +}_{\text{10}}^{\text{20}}\text{He}_{\text{2}}^{\text{4}}\\ \text{Atomic masses are given to be ,}\\ \text{m}\left(\text{H}_{\text{1}}^{\text{1}}\right)\text{= 1.007825 u, m}\left(\text{H}_{\text{1}}^{\text{2}}\right)\text{= 2.014102 u, m}\left(\text{H}_{\text{1}}^{\text{3}}\right)\text{= 3.016049 u,}\\ \text{m}\left(\text{C}_{\text{6}}^{\text{12}}\right)\text{= 12.00000 u, m}\left(\text{Ne}_{\text{10}}^{\text{20}}\right)\text{= 19.992439 u, m}\left(\text{He}_{\text{2}}^{\text{4}}\right)\text{= 4.002603 u\hspace{0.17em}}\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{The given reaction is}\\ \mathrm{H}_1^1+\mathrm{H}_1^3-\to \mathrm{H}_1^2+\mathrm{H}_1^2\\ \text{Q}=\left[{\text{m}}_{\text{N}}\left(\mathrm{H}_1^1\right)+{\text{m}}_{\text{N}}\left(\mathrm{H}_1^3\right)–{\text{m}}_{\text{N}}\left(\mathrm{H}_1^2\right)–{\text{m}}_{\text{N}}\left(\mathrm{H}_1^2\right)\right]\times 931\mathrm{MeV}\dots \dots \dots \dots .\left(\text{i}\right)\\ {\text{where m}}_{\text{N}}\text{\hspace{0.33em}refers to nuclear masses},\\ \text{If m refers to atomic masses},\text{then}\\ \text{m}\left(\mathrm{H}_1^1\right)={\text{m}}_{\text{N}}\left(\mathrm{H}_1^1\right)+{\text{m}}_{\text{e}}\\ \to {\text{m}}_{\text{N}}\left(\mathrm{H}_1^1\right)=\text{m}\left(\mathrm{H}_1^1\right)–{\text{m}}_{\text{e}}\\ \text{Similarly},\\ {\text{m}}_{\text{N}}\left(\mathrm{H}_1^3\right)=\text{m}\left(\mathrm{H}_1^3\right)–{\text{m}}_{\text{e}}\text{and}\\ {\text{m}}_{\text{N}}\left(\mathrm{H}_2^1\right)=\text{m}\left(\mathrm{H}_2^1\right)-{\text{m}}_{\text{e}}\\ \text{Thus, Q}=\left[{\text{m}}_{\text{N}}\left(\mathrm{H}_1^1\right)+{\text{m}}_{\text{N}}\left(\mathrm{H}_1^3\right)–{\text{m}}_{\text{N}}\left(\mathrm{H}_1^2\right)–{\text{m}}_{\text{N}}\left(\mathrm{H}_1^2\right)\right]\times 931\mathrm{MeV}\\ \text{Becomes Q}=\left[\text{m}\left(\mathrm{H}_1^1\right)+\text{m}\left(\mathrm{H}_1^3\right)-\text{m}\left(\mathrm{H}_1^2\right)–\text{m}\left(\mathrm{H}_1^2\right)\right]\times 931\mathrm{MeV}\\ \text{Q}=\left[\text{1}.00\text{7825}+\text{3}.0\text{16}0\text{49}–\text{2}.0\text{141}0\text{2}–\text{2}.0\text{141}0\text{2}\right]\times 931\mathrm{MeV}\\ \text{Q}=-0.00\text{433}\times \text{931}\\ \text{Q}=-\text{4}.0\text{3 MeV\hspace{0.17em}}.\\ \text{Since Q value is negative},\text{the reaction is endothermic}.\end{array}$

$\begin{array}{l}\left(\text{ii}\right)\text{\hspace{0.33em}The given reaction is}\\ \mathrm{C}_6^{12}+\mathrm{C}_6^{12}-\to \mathrm{Ne}_{10}^{20}+\mathrm{He}_2^4\\ \text{Q}=[{\text{m}}_{\text{N}}\left(\mathrm{C}_6^{12}+{\text{m}}_{\text{N}}\left(\mathrm{C}_6^{12}\right)-{\text{m}}_{\text{N}}\left(\mathrm{Ne}_{10}^{20}\right)-{\text{m}}_{\text{N}}\left(\mathrm{He}_2^4\right)\right]\times \text{931\hspace{0.17em}MeV}\dots \dots .\left(\text{2}\right)\\ \text{Where}{\text{\hspace{0.33em}m}}_{\text{N}}\text{\hspace{0.33em}refers to the nuclear mass}.\text{If m refers to the}\\ \text{atomic mass},\text{then}\\ {\text{m}}_{\text{N}}\left(\mathrm{C}_6^{12}\right)=\text{m}\left(\mathrm{C}_6^{12}\right)-{\text{6m}}_{\text{e}},\text{As\hspace{0.33em}}\mathrm{C}_6^{12}\text{has 6 electrons in its atom}.\\ \text{Similarly}\\ {\text{m}}_{\text{N}}\left(\mathrm{Ne}_{10}^{20}\right)=\text{m}\left(\mathrm{Ne}_{10}^{20}\right)-\text{1}0{\text{m}}_{\text{e}}\mathrm{and}{\text{m}}_{\text{N}}\left(\mathrm{He}_2^4\right)=\text{m}\left(\mathrm{He}_2^4\right)-{\text{2m}}_{\text{e}}\\ \text{Hence,}\\ \text{Q}=\left[{\text{m}}_{\text{N}}\left(\mathrm{C}_6^{12}\right)+{\text{m}}_{\text{N}}\left(\mathrm{C}_6^{12}\right)-{\text{m}}_{\text{N}}\left(\mathrm{Ne}_{10}^{20}\right)-{\text{m}}_{\text{N}}\left(\mathrm{He}_2^4\right)\right]\times 931\mathrm{MeV}\\ \text{Q}=\left[\begin{array}{l}\text{m}\left(\mathrm{C}_6^{12}\right)-{\text{6m}}_{\text{e}}+\text{m}\left(\mathrm{C}_6^{12}\right)-{\text{6m}}_{\text{e}}-\text{m}\left(\mathrm{Ne}_{10}^{20}\right)\\ +\text{1}0{\text{m}}_{\text{e}}-\text{m}\left(\mathrm{He}_2^4\right)+{\text{2m}}_{\text{e}}\end{array}\right]\times 931\mathrm{MeV}\\ \text{Q}=\left[\text{m}\left(\mathrm{C}_6^{12}\right)+\text{m}\left(\mathrm{C}_6^{12}\right)-\text{m}\left(\mathrm{Ne}_{10}^{20}\right)-\text{m}\left(\mathrm{He}_2^4\right)\right]\times 931\mathrm{MeV}\\ \text{Q}=\left[\text{12}.000000+\text{12}.000000–\text{19}.\text{992439}–\text{4}.00\text{26}0\text{3}\right]\times 931\mathrm{MeV}\\ \text{Q}=0.00\text{4958}\times \text{931 MeV}\\ \text{Q}=\text{4}.\text{66 MeV}\\ \text{Since Q value is positive},\text{the reaction is exothermic\hspace{0.17em}}.\end{array}$

Q.16

$\begin{array}{l}\text{Suppose, we think of fission of a}\text{Fe\hspace{0.33em}nucleus into two equal}_{\text{26}}^{\text{56}}\\ \text{fragments,\hspace{0.33em}}\text{Al. Is the fission energetically possible? Argue}_{\text{13}}^{\text{28}}\\ \text{by working out Q of the process. Given:}\\ \text{m}\left(\text{Fe}_{\text{26}}^{\text{56}}\right)\text{= 55.93494 u; m}\left(\text{Al}_{\text{13}}^{\text{28}}\right)\text{= 27.98191 u}\end{array}$

Ans.

$\begin{array}{l}\text{As suggested in the question},\text{if\hspace{0.33em}}\mathrm{Fe}_{26}^{56}\to \mathrm{Al}_{13}^{28}+\mathrm{Al}_{13}^{28}\\ \text{Then Q}-\text{value of process will be}\\ =\left[\text{m}\left(\mathrm{Fe}_{26}^{56}\right)–\text{2m}\left(\mathrm{Al}_{13}^{28}\right)\right]\times 931\mathrm{MeV}\\ =\left[\text{55}.\text{93494}–\text{2}\left(\text{27}.\text{98191}\right)\right]\times 931\mathrm{MeV}\\ =-0.0\text{2888}\times 931\mathrm{MeV}\\ =-0.0\text{2888}\times \text{931\hspace{0.17em}}\mathrm{MeV}\\ \text{Hence,}\\ \text{Q}=-\text{26}.\text{9}0\text{MeV\hspace{0.17em}}.\\ \text{Since, Q\hspace{0.33em}value is negative},\text{so the fission is not possible\hspace{0.17em}}.\end{array}$

Q.17

$\begin{array}{l}\text{The fission properties of}\text{Pu\hspace{0.33em}are very similar to those of\hspace{0.33em}}_{\text{94}}^{\text{239}}\text{U.}_{\text{92}}^{\text{235}}\\ \text{The average energy released per fission is 180 MeV. How much}\\ \text{energy, in MeV, is released if all atoms in 1 kg of pure}\text{Pu}_{\text{94}}^{\text{239}}\\ {\text{undergo fission?The fission properties of}}_{\text{94}}{\text{Pu}}^{\text{239}}\text{are very similar}\\ \text{to those of}\text{U. The average energy released per fission is 180 MeV.}_{\text{92}}^{\text{235}}\\ \text{How much energy, in MeV, is released if all atoms in 1 kg of}\\ \text{pure}\text{Pu\hspace{0.33em}undergo fission?}_{\text{94}}^{\text{239}}\end{array}$

Ans.

$\begin{array}{l}\text{Number of atoms present in 239 g of}\text{Pu= 6.023×10}_{\text{23}}^{\text{94}}{}_{\text{239}}\\ \text{Number of atoms present in 1 g of \hspace{0.33em}}\text{Pu\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.17em}=}_{\text{94}}^{\text{239}}\frac{{\text{6.023×10}}^{\text{23}}}{\text{239}}\\ \text{Number of atoms present in 1000 g of \hspace{0.33em}}\text{Pu\hspace{0.17em}=}_{\text{94}}^{\text{239}}\frac{{\text{6.023×10}}^{\text{23}}\text{×1000}}{\text{239}}\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= 2.52×10}}^{\text{24}}\text{atoms}\\ \text{As given,}\\ \text{the average energy released per fission = 180 MeV}\\ {\text{Thus total energy released\hspace{0.17em},\hspace{0.33em}Q=180 x 2.52 x 10}}^{\text{24}}\text{\hspace{0.17em}MeV}\\ {\text{Q=4.536×10}}^{\text{26}}\text{\hspace{0.17em}MeV}\\ {\text{Hence,\hspace{0.33em}4.536×10}}^{\text{26}}\text{MeV\hspace{0.33em}is\hspace{0.33em}released\hspace{0.33em}if\hspace{0.33em}all\hspace{0.33em}the\hspace{0.33em}atoms\hspace{0.33em}in\hspace{0.33em}1\hspace{0.17em}kg\hspace{0.33em}of\hspace{0.33em}pure}\\ \text{Pu\hspace{0.33em}undergo\hspace{0.33em}fission.}_{\text{94}}^{\text{239}}\end{array}$

Q.18

$\begin{array}{l}\text{A 1000 MW fission reactor consumes half of its fuel in 5.00 y.}\\ \text{How much}\text{U\hspace{0.33em}did it contain initially? Assume that the reactor}_{\text{92}}^{\text{235}}\\ \text{operates 80 \% of the time, that all the energy generated arises}\\ \text{from the fission of}\text{U\hspace{0.33em}and that this nuclide is consumed by}_{\text{92}}^{\text{235}}\\ \text{the fission process.}\end{array}$

Ans.

$\begin{array}{l}\text{Power of the fission reactor}\\ \text{P=1000 MW}\\ {\text{=1000×10}}^{\text{6}}\text{W}\\ {\text{=10}}^{\text{9}}\text{W}\\ \text{As given the time taken to consume\hspace{0.33em}half of the fuel is ,\hspace{0.33em}}\\ \text{t=5\hspace{0.33em}year}\\ \text{=5×365×24×60×60}\\ {\text{=1.577×10}}^{\text{8}}\text{s}\\ \text{Thus, energy consumed during this time of 5 years will be}\\ \text{E=P\hspace{0.17em}t}\\ {\text{E=10}}^{\text{9}}{\text{×1.577×10}}^{\text{8}}\\ {\text{=1.577×10}}^{\text{17}}\text{\hspace{0.17em}J}\\ \text{As we know energy generated per fission of}\text{U=200\hspace{0.17em}MeV}_{\text{92}}^{\text{235}}\\ {\text{=200×1.6×10}}^{\text{-13}}\text{J}\\ {\text{=3.2×10}}^{\text{-11}}\text{J}\\ \text{Therefore number of total fissions that have occurred in five years.}\\ \text{=}\frac{{\text{1.577×10}}^{\text{17}}}{{\text{3.2×10}}^{\text{-11}}}\\ {\text{=4.93×10}}^{\text{27}}\\ {\text{Now \hspace{0.33em}6.023×10}}^{\text{23}}\text{\hspace{0.33em}atoms}\left(\text{fissions}\right)\text{are produced by 235 g of}\text{U}_{\text{92}}^{\text{235}}\\ \text{Mass of\hspace{0.33em}}\text{U\hspace{0.33em}consumed in five years}_{\text{92}}^{\text{235}}\\ \text{=}\frac{{\text{235×4.93×10}}^{\text{27}}\text{g}}{{\text{6.023×10}}^{\text{23}}}\\ \text{= 1924 kg}\\ \text{=half of the mass of the fuel}\\ \text{consumed during 5 years as given in the question.}\\ \text{Hence, initial mass of\hspace{0.33em}}\text{U=2×1924}_{\text{92}}^{\text{235}}\\ \text{=3848\hspace{0.17em}kg}\end{array}$

Q.19

\begin{array}{l}\text{How long an electric lamp of 100 W can be kept glowing by fusion of 2}\text{.0 kg of deuterium?}\\ \text{The fusion reaction can be taken as}\\ \text{H}_{\text{1}}^{\text{2}}\text{+}\text{H}_{\text{1}}^{\text{2}}\stackrel{}{\to }\text{H}_{\text{2}}^{\text{3}}\text{+ n + 3}\text{.27 MeV}\end{array}

Ans.

\begin{array}{l}\text{As given power of lamp, P= 100 W and}\text{mass of deuterium= 2 kg}\\ \text{From the reaction given in the exercise, we find that 2 atoms of}\text{H}_{\text{1}}^{\text{2}}\\ \text{combine to give 3}\text{.2 MeV of energy}\text{.}\\ \text{Now 2}\text{g of}\text{H}_{\text{1}}^{\text{2}}\text{equivalent to 6}{\text{.023×10}}^{\text{23}}\text{atoms}\\ \text{Thus 2}\text{kg or 2000}\text{g of}\text{H}_{\text{1}}^{\text{2}}\text{is equivalent to}\\ \frac{\text{6}{\text{.023×10}}^{\text{23}}\text{×2000}}{\text{2}}{\text{=6023×10}}^{\text{23}}\\ \text{=6}{\text{.023 ×10}}^{\text{26}}\text{atoms}\\ \text{Thus the energy released for combination of 6}{\text{.023×10}}^{\text{26}}\text{atoms}\\ \text{should be}\\ \text{E=}\frac{\text{3}\text{.27×6}{\text{.023×10}}^{\text{26}}}{\text{2}}\\ \text{= 9}{\text{.85×10}}^{\text{26}}\text{MeV}\\ \text{E=9}{\text{.85×10}}^{\text{26}}\text{×1}{\text{.6×10}}^{\text{-13}}\text{J}\\ \text{E=1}{\text{.58×10}}^{\text{14}}\text{J}\\ \text{As}\text{P=}\frac{\text{E}}{\text{t}}\text{thus t=}\frac{\text{E}}{\text{P}}\text{and we get}\\ \text{t=}\frac{\text{1}{\text{.58×10}}^{\text{14}}}{\text{100}}\\ \text{= 1}{\text{.58×10}}^{\text{12}}\text{s}\\ \text{=}\frac{\text{1}{\text{.58×10}}^{\text{12}}}{\text{365×24×60×60}}\text{years}\end{array}

Q.20 Calculate the height of the potential barrier for the head-on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

Ans.

\begin{array}{l}\text{Before collision},\text{the initial total energy E of the two deuterons}\\ \text{is given by,}\\ \text{E}=\text{2}\left(\text{Kinetic energy}\right)\\ \text{After collision},\text{when the two deuterons stop},\text{their energy is totally}\\ \text{potential energy U and it is given by}\\ \text{U}=\frac{\text{1}}{\text{4}\pi {Є}_0}\frac{\text{e}.\text{e}}{\text{2}\text{R}}\\ =\frac{\text{1}}{\text{4}\pi {Є}_0}\frac{{\text{e}}^{\text{2}}}{\text{2}\text{R}}\\ \text{As given that radius of each deuteron sphere is}\text{R}=\text{2 fm}\\ =\text{2 x 1}{0}^{-\text{15}}\text{m}\\ \text{Charge,}\text{e}=\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\text{C}\text{and}\frac{\text{1}}{\text{4}\pi {Є}_0}=\text{9}\times \text{1}{0}^{\text{9}}{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\\ Thus\text{P}.\text{E}=\frac{\left[\text{9}\times \text{1}{0}^{\text{9}}\times {\left(\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\right)}^{\text{2}}\right]}{\left(\text{4}\times \text{1}{0}^{-\text{15}}\right)}joule\\ \text{P}.\text{E}=\frac{\text{9}\times \text{1}.\text{6}\times \text{1}.\text{6}\times \text{1}{0}^{-\text{14}}}{\text{4}\times 1.6\times \text{1}{0}^{-\text{19}}}eV\\ =3.6\times \text{1}{0}^{5}eV\\ =360keV\\ As\text{per law of conservation of energy},\text{we can say that}\\ \text{2}\text{K}\text{.E}=\text{P}\text{.E}\\ K.E=\frac{P.E}{2}\\ =\frac{360}{2}\\ =180\text{}keV.\end{array}

Q.21 From the relation R = R₀ A^1/3, where R₀ is a constant and A is the mass number of a nucleus, show that nuclear matter density is nearly constant (i.e independent of A)

Ans

\begin{array}{l}\text{As nuclear density}=\frac{\text{mass number}}{\text{volume}}\\ =\frac{\text{3}\text{A}}{\left(\text{4}\pi {\text{R}}^{\text{3}}\right)}\\ =\frac{\text{3}\text{A}}{\text{4}\pi \text{R}{}_0{}^{\text{3}}\text{A}}\\ =\frac{\text{3}}{\text{4}\pi {\text{R}}_0{}^{\text{3}}}\\ \text{By using relation}\text{R}={\text{R}}_0{\text{A}}^{\frac{1}{3}}\\ \text{Hence, nuclear density is independent of A}.\end{array}

Q.22

\begin{array}{l}\text{For the}\left(\text{positron}\right)\text{emission from a nucleus},\text{there is another competing}\text{process known as electron capture}\\ \left(\text{electron from an inner orbit},\text{say},\text{the K}-\text{shell},\text{is captured by the nucleus and a neutrino is emitted}\right).\\ {\text{e}}^{-}+X_Z^A---\to Y_{Z-1}^A+v\\ \text{Show that if}\text{emission is energetically allowed},\text{electron capture is necessarily allowed but not vice}-\text{versa}.\end{array}

Ans.

\begin{array}{l}\text{For the positron emission},\text{we have}\\ X_Z^A---\to Y_{Z-1}^A+\text{e}_1{}^0+v+{\text{Q}}_{\text{1}}\mathrm{\dots \dots }\left(\text{1}\right)\\ \text{And for electron capture},\text{we have}\\ X_Z^A+\text{e}_{-1}{}^0-----\to Y_{Z-1}^A+v+{\text{Q}}_{\text{2}}\mathrm{\dots \dots }\left(\text{2}\right)\\ {\text{If m}}_{\text{N}}\text{represents the nuclear mass and m represents the}\\ \text{atomic mass},{\text{then for m}}_{\text{e}}{\text{as the mass of}}_{-\text{1}}{\text{e}}^0\text{or}{}_{+\text{1}}{\text{e}}^0,\\ \text{we have nuclear mass of}\\ X_Z^A={\text{m}}_{\text{N}}\left(X_Z^A\right)\\ =\left[\text{m}\left(X_Z^A\right)–{\text{Zm}}_{\text{e}}\right]\\ \text{Similarly nuclear mass of}Y_{Z-1}^A\text{should be}\\ Y_{Z-1}^A={\text{m}}_{\text{N}}\left(Y_{Z-1}^A\right)\\ =\left[\text{m}\left(Y_{Z-1}^A\right)–\left(\text{Z}-\text{1}\right){\text{m}}_{\text{e}}\right]\\ Hencefromequation(1)\\ {\text{Q}}_{\text{1}}=\left[{\text{m}}_{\text{N}}\left(X_Z^A\right){\text{-m}}_{\text{N}}\left(Y_{Z-1}^A\right)–{\text{m}}_{\text{e}}\right]{\text{c}}^{\text{2}}\\ =\left[\text{m}\left(X_Z^A\right)–\text{Z}{\text{m}}_{\text{e}}\text{-m}\left(Y_{Z-1}^A\right)+\left(\text{Z}-\text{1}\right){\text{m}}_{\text{e}}–{\text{m}}_{\text{e}}\right]{\text{c}}^{\text{2}}\\ =\left[\text{m}\left(X_Z^A\right)–\text{m}\left(Y_{Z-1}^A\right)-{\text{2m}}_{\text{e}}\right]{\text{c}}^{\text{2}}\mathrm{\dots \dots ..}\left(\text{3}\right)\\ \text{Similarly from equaton}\left(\text{2}\right)\\ {\text{Q}}_{\text{2}}=\left[{\text{m}}_{\text{N}}\left(X_Z^A\right)+{\text{m}}_{\text{e}}–{\text{m}}_{\text{N}}\left(Y_{Z-1}^A\right)\right]{\text{c}}^{\text{2}}\\ =\left[\text{m}\left(X_Z^A\right){\text{– Zm}}_{\text{e}}+{\text{m}}_{\text{e}}–\text{m}\left(Y_{Z-1}^A\right)+\left(\text{Z}-\text{1}\right){\text{m}}_{\text{e}}\right]{\text{c}}^{\text{2}}\\ =\left[\text{m}\left(X_Z^A\right)–\text{m}\left(Y_{Z-1}^A\right)\right]{\text{c}}^{\text{2}}\dots \dots \dots \dots \left(\text{4}\right)\\ \text{Thus from eqns}.\left(\text{3}\right)\text{and}\left(\text{4}\right),\text{we find that}\\ {\text{if Q}}_{\text{1}}>0,\text{is necessarily greater than zero}.\\ \text{However},{\text{if Q}}_{\text{2}}>0,{\text{it does not always mean that Q}}_{\text{1}}>0.\end{array}

Q.23

$\begin{array}{l}\text{In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural}\\ \text{abundance on Earth.\hspace{0.17em}The three isotopes and their masses are \hspace{0.17em}}\text{Mg}_{\text{12}}^{\text{24}}\left(\text{23.98504 u}\right)\text{,}\text{Mg}_{\text{12}}^{\text{25}}\left(\text{24.98584 u}\right)\text{and}\text{Mg}_{\text{12}}^{\text{26}}\left(\text{25.98259 u}\right)\text{.\hspace{0.17em}}\\ \text{The natural abundance of}\text{Mg \hspace{0.17em}is 78.99 \% by mass. Calculate the abundance of the other two isotopes.}_{\text{12}}^{\text{24}}\end{array}$

Ans.

$\begin{array}{l}\text{Let us assume that the natural abundance of}\text{Mg\hspace{0.17em}be x\% then natural abundance of}_{\text{12}}^{\text{25}}\text{Mg\hspace{0.17em} will be}_{\text{12}}^{\text{26}}\left(\text{100 – x – 78.99}\right)\text{\%.}\\ \text{Thus,}\\ \text{24.312 =}\frac{\left[\text{78.99×23.98504 + 24.98584x +}\left(\text{100 – x – 78.99}\right)\left(\text{25.98259}\right)\right]}{\text{100}}\\ \text{24.312 =}\frac{\text{2440.474 – 0.99675\hspace{0.17em}x}}{\text{100}}\\ \text{0.99675\hspace{0.17em}x = 2440.479 – 2431.2}\\ \text{x =}\frac{\text{9.274}}{\text{0.99675}}\text{= 9.304\hspace{0.17em}\%}\\ \text{Thus natural abundance of\hspace{0.17em}}\text{Mg = 9.304\hspace{0.17em}\%\hspace{0.17em} and for\hspace{0.17em}}_{\text{12}}^{\text{25}}\text{Mg,\hspace{0.17em}(21.01 – x) = 11.71\hspace{0.17em}\%\hspace{0.17em}.}_{\text{12}}^{\text{26}}\end{array}$

Q.24

$\begin{array}{l}\text{The neutron separation energy is defined as the energy required}\\ \text{to remove a neutron from the nucleus. Obtain the neutron separation}\\ \text{energies of the nuclei\hspace{0.33em}}\text{Ca\hspace{0.33em}and}_{\text{20}}^{\text{41}}\text{Al\hspace{0.33em}from the following data:}_{\text{13}}^{\text{27}}\\ \text{m}\left(\text{Ca}_{\text{20}}^{\text{40}}\right)\text{=39.962591 u,}\\ \text{m}\left(\text{Ca}_{\text{20}}^{\text{41}}\right)\text{=40.962278 u}\\ \text{m}\left(\text{Al}_{\text{13}}^{\text{26}}\right)\text{=25.986895 u,}\\ \text{m}\left(\text{Al}_{\text{13}}^{\text{27}}\right)\text{=26.981541 u.}\end{array}$

Ans.

$\begin{array}{l}\text{The process of neutron separation for}\text{Ca,}_{\text{20}}^{\text{41}}\\ \text{is given as}\\ \text{Ca+E=}_{\text{20}}^{\text{41}}\text{Ca+}_{\text{20}}^{\text{40}}\text{n}_{\text{0}}^{\text{1}}\\ \text{E=}\left[\text{m}\left(\text{Ca}_{\text{20}}^{\text{40}}\right){\text{+m}}_{\text{n}}\text{–m}\left(\text{Ca}_{\text{20}}^{\text{41}}\right)\right]\text{×931\hspace{0.17em}MeV}\\ \text{E=}\left[\text{39.962591+1.008665–40.962278}\right]\text{×931\hspace{0.17em}MeV}\\ \text{E=0.008978×931=8.36 MeV}\\ \text{The process of neutron separation for\hspace{0.33em}}\text{Al\hspace{0.33em}is}_{\text{13}}^{\text{27}}\\ \text{Al+E=}_{\text{13}}^{\text{27}}\text{Al+}_{\text{13}}^{\text{26}}\text{n}_{\text{0}}^{\text{1}}\\ \text{Thus, neutron separation energy will be}\\ \text{E=}\left[\text{m}\left(\text{Al}_{\text{13}}^{\text{26}}\right){\text{+m}}_{\text{n}}\text{–m}\left(\text{Al}_{\text{13}}^{\text{27}}\right)\right]\text{×931\hspace{0.17em}MeV}\\ \text{E=}\left[\text{25.986895+1.008665-26.981541}\right]\text{×931\hspace{0.17em}MeV}\\ \text{Thus,\hspace{0.33em}E =0.014019×931}\\ \text{=13.05\hspace{0.17em}MeV\hspace{0.17em}.}\end{array}$

Q.25

\begin{array}{l}\text{A source contains two phosphorus radio nuclides}\text{P}_{\text{15}}^{\text{32}}\left({\text{T}}_{\frac{\text{1}}{\text{2}}}\text{= 14}\text{.3d}\right)\text{and}\\ \text{P}_{\text{15}}^{\text{33}}\left({\text{T}}_{\frac{\text{1}}{\text{2}}}\text{=25}\text{.3d}\right)\text{. Initially,10 \%}\text{of the decays come from}\text{P}_{\text{15}}^{\text{33}}\text{.}\\ \text{How long one must wait until 90 \% do so?}\end{array}

Ans.

\begin{array}{l}{\text{Let us assume that R}}_{\text{01}}{\text{and R}}_{\text{02}}\text{be the initial activities of}\text{P}_{\text{15}}^{\text{33}}\text{and}\text{P}_{\text{15}}^{\text{32}}\\ {\text{respectively and R}}_{\text{1}}{\text{and R}}_{\text{2}}\text{be their activities at any instant t}\text{.}\\ \text{As suggested in the question,}\\ {\text{R}}_{\text{01}}\text{=}\frac{\text{10}\left[{\text{R}}_{\text{01}}{\text{+ R}}_{\text{02}}\right]}{\text{100}}{\text{as R}}_{\text{01}}\text{=10\%}\text{of total activity}\text{.}\\ \text{Hence}\\ {\text{10R}}_{\text{01}}{\text{=R}}_{\text{01}}{\text{+ R}}_{\text{02}}\\ \text{Or}\\ {\text{R}}_{\text{02}}{\text{=9R}}_{\text{01}}\text{………}\text{.}\left(\text{1}\right)\\ \text{At a later time t ,}{\text{R}}_{\text{1}}\text{=}\frac{\text{90}\left[{\text{R}}_{\text{1}}{\text{+ R}}_{\text{2}}\right]}{\text{100}}\\ \text{as}{\text{R}}_{\text{1}}\text{= 90\%}\text{of total activity at time t}\text{.}\\ \text{Hence}\\ {\text{10R}}_{\text{1}}{\text{=9R}}_{\text{1}}{\text{+9R}}_{\text{2}}\\ \text{Or}\\ {\text{R}}_{\text{2}}\text{=}\frac{{\text{10 R}}_{\text{1}}{\text{–9R}}_{\text{1}}}{\text{9}}\text{=}\frac{{\text{R}}_{\text{1}}}{\text{9}}\text{……}\mathrm{..}\left(\text{2}\right)\\ \text{Now dividing}\left(\text{2}\right)\text{by}\left(\text{1}\right)\text{we get,}\\ \frac{{\text{R}}_{\text{2}}}{{\text{R}}_{\text{02}}}\text{=}\left(\frac{{\text{R}}_{\text{1}}}{\text{9}}\right)\left(\frac{\text{1}}{{\text{9R}}_{\text{01}}}\right)\\ \text{Or}\\ \frac{{\text{R}}_{\text{2}}}{{\text{R}}_{\text{02}}}\text{=}\frac{\text{1}}{\text{81}}\frac{{\text{R}}_{\text{1}}}{{\text{R}}_{\text{01}}}\\ \text{As per decay law,}{\text{R = R}}_{\text{0}}{\text{e}}^{\text{-λt}}\\ \frac{{\text{R}}_{\text{02}}{\text{e}}^{\text{-λ2t}}}{{\text{R}}_{\text{02}}}\text{=}\frac{\text{1}}{\text{81}}\frac{{\text{R}}_{\text{01}}{\text{e}}^{\text{-λ1t}}}{{\text{R}}_{\text{01}}}\\ \text{Þ 81=}\frac{{\text{e}}^{\text{-λ1t}}}{{\text{e}}^{\text{-λ2t}}}\\ \text{Or}\\ \text{e}\left({\text{λ}}_{\text{2}}{\text{– λ}}_{\text{1}}\right)\text{t =81}\\ \text{Taking natural log of both sides,}\\ \text{ln}\text{e}\left({\text{λ}}_{\text{2}}{\text{– λ}}_{\text{1}}\right)\text{t =ln 81}\\ \text{Or}\left({\text{λ}}_{\text{2}}{\text{– λ}}_{\text{1}}\right)\text{t=2}\text{.303 log 81 ……}\text{.}\left(\text{3}\right)\\ \text{As ,}\text{λ =}\frac{\text{0}\text{.693}}{{\text{T}}_{\frac{\text{1}}{\text{2}}}}\\ \text{Thus}{\text{λ}}_{\text{1}}\text{=}\frac{\text{0}\text{.693}}{{\left({\text{T}}_{\frac{\text{1}}{\text{2}}}\right)}_{\text{1}}}\text{=}\frac{\text{0}\text{.693}}{\text{25}\text{.3}}{\text{day}}^{\text{-1}}{\text{and λ}}_{\text{2}}\text{=}\frac{\text{0}\text{.693}}{{\left({\text{T}}_{\frac{\text{1}}{\text{2}}}\right)}_{\text{2}}}\\ \text{=}\frac{\text{0}\text{.693}}{\text{14}\text{.3}}{\text{day}}^{\text{-1}}\\ \text{Thus with reference to equation}\left(\text{3}\right)\text{, we get}\\ \left(\frac{\text{0}\text{.693}}{\text{14}\text{.3}}\text{–}\frac{\text{0}\text{.693}}{\text{25}\text{.3}}\right)\text{t=2}\text{.303 log 81}\\ \text{Or}\\ \frac{\text{0}\text{.693×}\left(\text{25}\text{.3-14}\text{.3}\right)\text{t}}{\text{25}\text{.3 ×14}\text{.3}}\text{= 2}\text{.303}\text{log 81}\\ \frac{\text{0}\text{.693×}\left(\text{11}\right)\text{t}}{\text{25}\text{.3 ×14}\text{.3}}\text{= 2}\text{.303}\text{log 81}\\ \text{Thus}\\ \text{t=}\left[\frac{\left(\text{2}\text{.303 log 81}\right)\left(\text{25}\text{.3×14}\text{.3}\right)}{\text{11×0}\text{.693}}\right]\\ \text{t=}\frac{\left[\text{2}\text{.303×1}\text{.9085×25}\text{.3×14}\text{.3}\right]}{\left[\text{11×0}\text{.693}\right]}\\ \text{t =208}\text{. 6 days»209 days}\text{.}\end{array}

Q.26

\begin{array}{l}\text{Under certain circumstances, a nucleus can decay by emitting a}\text{particle}\\ \text{more massive than an–particle}\text{. Consider the following decay process:}\\ R_{88}^{223}a--\to P_{82}^{209}b+C_6^{14}\\ R_{88}^{223}a---\to R_{86}^{219}n+H_2^{4}e\\ \text{Calculate the Q values for these decays and determine that the both are}\\ \text{energetically allowed}\text{.The required data is}\\ \text{m}\left(R_{88}^{223}a\right)=\text{223}.0\text{185}0\text{u,}\text{m}\left(P_{82}^{209}b\right)=\text{2}0\text{8}.\text{981}0\text{7 u,}\\ \text{m}\left(R_{86}^{219}n\right)=\text{219}.00\text{948 u,}\text{m}\left(C_6^{14}\right)=\text{14}.00\text{324 u,}\\ \text{m}\left(H_2^{4}e\right)=\text{4}.00\text{26}0\text{u}\text{.}\end{array}