# Nuclei

## Physics

### NCERT

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 $_5^{11}B$ .

(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: $m(_{26}^{56}Fe)=55.934939 u^m(_{83}^{209} B_i)=208.980388u$