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.