1   (a) Two stable isotopes of lithium $_3^6 Li$ and $_3^7 Li$ have respective abundances of $7.5$% and $92.5$%. These isotopes have masses $6.01512 u$ and $7.01600 u$, respectively. Find the atomic mass of lithium.$\\$ (b) Boron has two stable isotopes $_5^{10 }B$ and $_5^{10 }B$ . Their respective masses are $10.01294 u$ and $11.00931 u$, and the atomic mass of boron is $10.811 u$. Find the abundances of $_5^{10}B$ and $_5^{11}B$ .

Solution :

(a)Mass of $_3^6 Li$ lithium isotope,$m_1=6.01512 u$$\\$ Mass of $_3^7 Li$ lithium isotope ,$m=7.01600 u$$\\$ Abundance of $_3^6Li,n_1=7.5$%$\\$ Abundance of$_3^7Li,n_2=92.5$%$\\$ The atomic mass of lithium atom is given as:$\\$ $m=\dfrac{m_1n_1+m_2n_2}{n_1+n_2}\\ =\dfrac{6.01512*7.5+7.01600*92.5}{92.5+7.5}\\ =6.940934 u.$$\\$ (b)Mass of $_5^{10}B$ boron isotope $_5^{10}B,m=10.01294 u$$\\$ Mass of boron isotope $_5^{10} B,m=11.00931 u$$\\$ Abundance of $_5^{10}B,n_1=x $%$\\$ Abundance of $_5^{11}B,n_2=(100-x)$%$\\$ Atomic mass of boron, $m = 10.811 u$$\\$ The atomic mass of boron atom is given as:$m=\dfrac{m_1n_1+m_2n_2}{n_1+n_2}$$\\$ $10.811=\dfrac{10.01294*x+11.00931*(100-x)}{x+100-x}\\ 1081.11 = 10.01294 x+ 1100.931 -11.00931 x\\ \therefore x=\dfrac{19.821}{0.99637}=19.89$%$\\$ And $ 100-x=80.11$%$\\$ Hence, the abundance of is $_5^{10} B 19.89$% and that of $_5^{11}B$ is $ 80.11$%

2   The three stable isotopes of neon:$_{10}^{20}Ne,_{10}^{20}Ne$ and $_{10}^{21}Ne$ and $_{10}^{22}Ne$ have respective abundances of $90.51$%,$0.27$ % and $9.22 $ %.The atomic masses of the three isotopes are $19.99 u,20.99 u $ and $ 21.99 u,$ respectively. Obtain the average atomic mass of neon.

Solution :

Atomic mass of $_{10}^{20} Ne,m_1=19.99 u $$\\$ Abundance of $ _{10}^{20}Ne,\eta_1=90.51 $ %$\\$ Atomic mass of $_{10}^{21}Ne,m2=20.99 u$$\\$ Abundance of $_{10}^{21}Ne,\eta_2=0.27$%$\\$ Atomic mass of $_{10}^{22}Ne,m_3=21.99 u$$\\$ Abundance 0f $_{10}^{22}Ne,\eta_3=9.22 $%$\\$ The average atomic mass of neon is given as:$\\$ $m=\dfrac{m_1\eta_1+m_2\eta_2+m_3\eta_3}{\eta_1+\eta_2+\eta_3}\\ =\dfrac{19.99*90.51+20.99*0.27+21.99*9.22}{90.51+0.27+9.22}\\ =20.1771 u$

3   Obtain the binding energy (in MeV) of a nitrogen nucleus $(_7^{14}N)$,given $m(_7^{14}N)=14.00307 u$

Solution :

Atomic mass of$(_7N^{14})$ , given $m =14.00307 u$ A nucleus of $_7 N^{14}$ nitrogen contains $7$ protons and $7$ neutrons.$\\$ Hence, the mass defect of this nucleus, $\Delta m=7 m_H + 7 m_n - m$$\\$ Where,$\\$ Mass of a proton, $m_H = 1.007825 u$$\\$ Mass of a neutron, $m_n = 1.008665 u$$\\$ $\Delta m = 7 * 1.007825 + 7 * 1.008665 - 14.00307\\ = 7.054775 + 7.06055 - 14.00307\\ = 0.11236 u$$\\$ But $1 u = 931.5 MeV / c^2$$\\$ $\Delta m =0.11236 * 931.5 MeV / c^2$$\\$ Hence, the binding energy of the nucleus is given as: $E_b =\Delta mc^2$$\\$ Where, c = Speed of light$\\$ $E_b=0.11236*931.5(\dfrac {Me V})(c^2)*c^2\\ =104.66334 Me V$$\\$ Hence, the binding energy of a nitrogen nucleus is $104.66334 Me V.$

4   Obtain the binding energy of the nuclei $_{26}^{56} Fe$ and $_{83}^{209}B_i$ in units of MeV from the following data: $m(_{26}^{56}Fe)=55.934939 u^m(_{83}^{209} B_i)=208.980388u$

Solution :

Atomic mass of $_{26}^{56}Fe,m_1=55.934939 u$$\\$ $_{26}^{56}Fe$ nucleus has $26$ protons and $(56 - 26) = 30$ neutrons$\\$ Hence, the mass defect of the nucleus, $\Delta m =26 * m_H +30 * m_n - m_1$$\\$ Where,$\\$ Mass of a proton, $m_H = 1.007825 u$$\\$ Mass of a neutron,$ m_n = 1.008665 u$$\\$ $\Delta m=26*1.007825+30*1.008665-55.934939\\ =26.20345+30.25995-55.934939\\ =0.528461 u$$\\$ But $1 u=931.5 MeV/c62$$\\$ $\Delta m=0.528461*931.5 MeV/c^2$$\\$ The binding energy of this nucleus is given as:$\\$ $E_{b1}=\Delta mc^2$$\\$ Where,c= Speed of light$\\$ $\therefore E_{b1}=0.58461*931.5(\dfrac{MeV}{c^2})*c^2\\ =492.26MeV$$\\$ $\therefore E_{b1}=0.528461*931.5(\dfrac{MeV}{c^2})*c^2\\ =492.26 MeV$$\\$ Average binding energy per nucleon $=\dfrac{492.26}{56}=8.79 MeV$$\\$ Atomic mass of $_{83}^{209}B_i,m_2=208.980388 u$$\\$ $_{83}^{209}B_i$ nucleus has $83$ protons and $(209-83) 126$ neutrons.$\\$ Hence, the mass defect of this nucleus is given as:$\\$ $\Delta m'=83*m_H+126*m_n-m_2$$\\$ Where,$\\$ Mass of a proton,$m_H=1.007825 u$$\\$ Mass of a neutron, $m_n=1.008665 u$$\\$ $\Delta m'=83*1.007825+126*1.008665-208.980388\\ =83.649475+127.091790-208.980388\\ =1.760877 u$$\\$ But$1u=931.5 MeV/c^2$$\\$ $\Delta m'=1.760877*931.5 MeV/c^2$$\\$ Hence, the binding energy of this nucleus is given as:$\\$ $E_{b2}=\Delta m'c^2\\ =1.760877*931.5(\dfrac{MeV}{c^2})*c^2\\ =1640.26 MeV$$\\$ Average binding energy per nucleon $=\dfrac{1640.26}{209}=7.848 MeV$

5   A given coin has a mass of $3.0 g$. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of $_{29}^{63}Cu$ atoms (of mass $62.92960 u$).

Solution :

Mass of a copper coin, $m'=3g$$\\$ Atomic mass of $_{29}Cu^{63}$ atom, $ m=62.92960 u$$\\$ The total number of $_{29}Cu^{63} $ atoms in the coin\\ $,N=\dfrac{N_A*m'}{\text{Mass number}}$$\\$ Where,$\\$ $N_A=$ Avogadro’s number $=6.023* 10^{23} $ atoms/g$\\$ Mass number $= 63 g$$\\$ $\therefore N=\dfrac{6.023*10^{23}*3}{63}=2.868*10^{22}$$\\$ $_{29}Cu^{63}$ nucleus has $29$ protons and $(63-29) 34$ neutrons$\\$ $\therefore $ Mass defect of this nucleus,$ \Delta m'=29*m_H+34*m_n-m$$\\$ Where, $\\$ Mass of a proton,$ m_H=1.007825 u$$\\$ Mass of a neutron, $m_n = 1.008665 u$$\\$ $\therefore m ' = 29 * 1.007825 + 34 * 1.008665 - 62.9296\\ = 0.591935 u$$\\$ Mass defect of all the atoms present in the coin, $\Delta m = 0.591935 × 2.868 × 1022$$\\4 $=1.6976695*10^{22}u$$\\$ But$1u=931.5 MeV/c^2$$\\4 $\therefore \Delta m = 1.69766958 * 10^{22} * 931.5 MeV / c^2$$\\$ Hence, the binding energy of the nuclei of the coin is given as:$\\$ $E_b =\Delta mc^2\\ =1.69766958*10^{22}*931.5(\dfrac{MeV}{c^2})*c^2$$\\$ $=1.581* 10^{25}MeV$$\\$ But $1 MeV=1.6* 10^{-13}J\\ E_b=1.581*10^{25}*1.6*10^{-13}\\ =2.5296*10^{12}J$$\\$ This much energy is required to separate all the neutrons and protons from the given coin.

6   Write nuclear reaction equations for$\\$ (i)$\alpha -$ decay of $ _{88}^{226} Ra$$\\$ (ii)$ \alpha -$decay of $_{94}^{242}Pu$$\\$ (iii)$\beta -$decay of $_{15}^{32}P$$\\$ (iv)$\beta - $ dacay of $_{83}^{210}B_i$$\\$ (v)$\beta^+-$ decay of $_6^{11}C$$\\$ (vi)Electron capture of $_{54}^{120}Xe$$\\$

Solution :

$\alpha $ is a nucleus of helium $(_2He^ 4 )$ and $\beta $ is an electron $(e ^{- } \text{for} \beta ^- \text{and} e^+\text{ for} \beta +)$ In every $\alpha$ decay, there is a loss of $2$ protons and $2$ neutrons. In every $\beta ^+$-decay, there is a loss of $1$ proton and a neutrino is emitted from the nucleus. In every $\beta ^-$ -decay, there is a gain of $ 1$ proton and an antineutrino is emitted from the nucleus.$\\$ For the given cases, the various nuclear reactions can be written as:$\\$ (i)$_{88}Ra^{226}\to _{86}Rn^{222}+_2 He^4$$\\$ (ii)$_{94}^{242}Pu \to _{92}^{238}U+_{2}{4}He$$\\$ (iii)$_{15}^{32}P\to _{16}^{32}S+e^-+\bar{v}$$\\$ (iv)$_{83}^{210}B\to _{84}^{210}PO+e^-+\bar{v}$$\\$ (v)$_{6}^{11}C\to _5^{11}B+e^++v$$\\$ (vi)$_{43}^{97}Tc\to _{42}^{97}MO+e^++v$$\\$ (vii)$_{54}^{120}Xe+e^+\to _{53}^{120}I+v$$\\$

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

Solution :

Half-life of the radioactive isotope = T years$\\$ Original amount of the radioactive isotope = $N_ 0$$\\$ (a) After decay, the amount of the radioactive isotope = N$\\$ It is given that only $3.125\% $ of $N _0$ remains after decay. Hence, we can write:$\\$ $\dfrac{N}{N_0}=3.125%=\dfrac{3.125}{100}=\dfrac{1}{32}$$\\$ But$ \dfrac{N}{N_0}=e^{-\lambda I}$$\\$ Where,$\lambda $=Decay constant $t=$Time$\\$ $\therefore -\lambda t=\dfrac{1}{32}\\ -\lambda t=In l-In 32\\ -\lambda t=0-3.4657\\ t=\dfrac{3.4657}{\lambda}$$\\$ Since $\lambda =\dfrac{0.693}{T}\\ \therefore t=\dfrac{3.466}{\dfrac{0.693}{T}}\approx 5T \text{Years}$$\\$ Hence, the isotope will take about $5T$ years to reduce to $3.125\%$ of its original value.

(b) After decay, the amount of the radioactive isotope = N$\\$ It is given that only $1\%$ of N 0 remains after decay. Hence, we can write:$\\$ $\dfrac{N}{N_0}=1%=\dfrac{1}{100}$$\\$ But$ \dfrac{N}{N_0}=e^{-\lambda t}\\ \therefore e^{-\lambda t}=\dfrac{1}{100}\\ -\lambda t=In 1-In 100\\ -\lambda t =0-4.6052\\ t=\dfrac{4.6052}{\lambda}$$\\$ Since,$\lambda=0.639/T\\ \therefore t=\dfrac{4.6052}{\dfrac{0.693}{T}}=6.645 T $ years$\\$ Hence, the isotope will take about $6.645T$ years to reduce to $1\%$ of its original value.

8   The normal activity of living carbon-containing matter is found to be about $15$ decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive $_6^{14} C$ present with the stable carbon isotope $_6^{12}C$ . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ($5730$ years) of $_6^{14} C$ , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of $_6^{14} C$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Solution :

Decay rate of living carbon-containing matter, $ R = 15$ decay/min Let N be the number of radioactive atoms present in a normal carbon- containing matter.$\\$ Half life of $_6^{14}C,T_{1/2}=5730$ years$\\$ The decay rate of the specimen obtained from the Mohenjodaro site:$\\$ $R' = 9$ decays/min$\\$ Let $N'$ be the number of radioactive atoms present in the specimen during the Mohenjodaro period.$\\$ Therefore, we can relate the decay constant, $\lambda $ and time, $t$ as:$\\$ $\dfrac{N'}{N}=\dfrac{R'}{R}=e^{-\lambda t}\\ e^{-\lambda t}=\dfrac{9}{15}=\dfrac{3}{5}\\ -\lambda t =\log_e \dfrac{3}{5}=-0.5108\\ \therefore t=\dfrac{0.5108}{\lambda}\\ \text{But} \lambda =\dfrac{0.693}{T_{1/2}}\\ =\dfrac{0.693}{5730}\\ \therefore t=\dfrac{0.5108}{\dfrac{0.693}{5730}}\\ =4223.5 $ years$\\$ Hence, the approximate age of the Indus-Valley civilization is $4223.5 $years.

9   Obtain the amount of $ \ _{27}^{60}Co \ $ necessary to provide a radioactive source of $ 8.0mCi$ strength. The half-life of $_{27}^{60}Co $ is $ 5.3$ years.$\\$

Solution :

The strength of the radioactive source is given as:$\\$ $\dfrac{dN}{dt}=8.0mCi\\ =8*10^{-3}*3.7*10^{10}\\ =29.6*10^{7}\text{decay}/s$$\\$ Where,$\\$ N=Required number of atoms$\\$ Half-life of $_{27}^{60}Co,T_{1/2}=5.3 $years$\\$ $=5.3*365*24*60*60\\ =1.67*108 s$$\\$ For decay constant $\lambda $ , we have the rate of decay as: $\\$ $\dfrac{dN}{dt}=\lambda N\\ \text{Where}\\ \lambda =\dfrac{0.693}{T_{1/2}}=\dfrac{0.693}{1.67810^8}s^{-1}\\ \therefore N=\dfrac{1}{\lambda}\dfrac{dN}{dt}\\ =\dfrac{29.6*10^7}{\dfrac{0.693}{1.67*10^8}}=7.133*10^{16} \text{atoms}$$\\$ For $_27Co^{60}$$\\$ Mass of $6.023 × 1023$ (Avogadro’s number) atoms = $60 g$$\\$ Mass of $7.133 * 10 atoms =\dfrac{60*7.133*10^{16}}{6.023*10^{23}}=7.106*10^{-6}g$$\\$ Hence, the amount of$_27Co^{60}$necessary for the purpose is $7.106 * 10 ^{-6} g .$

10   The half-life of $_{38}^{90}Sr $ is $ 28 $ years. What is the disintegration rate of $ 15 mg $ of this isotope?

Solution :

Half life of $_{38}^{90}Sr,t_{1/2}=28 $ years$\\$ $=28*365*24*60*60\\ =8.83*10^8 s$$\\$ Mass of the isotope, $m = 15 mg$$\\$ $90 g $ of $_{38}^{90}Sr$ atom contains $ 6.023*10^{23}$ (A vogadro's number)atoms$\\$ Therefore,$15mg $ 0f $_{38}^{90}Sr $ contains:$\\$ $\dfrac{6.023*10^{23}*15 * 10^{-3}}{90}$ ,i.e.$ 1.0038*10^{20}$ Number of atomms$\\$ Rate of disntegration, $ \dfrac{dN}{dt}=\lambda N$$\\$ Where,$\\$ $\lambda=$ decay constant=$ \dfrac{0.693}{8.83*10^{8}}s^{-1}$$\\$ $\therefore \dfrac{dN}{dt}=\dfrac{0.693*1.0038*10^{20}}{8.83*10^8}\\ =7.878*10^{10} $ atoms/s$\\$ Hence, the disintegration rate of $15 mg$ of the given isotope is $\\$ $7.878*10^{10}$ atoms/s