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

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