Question

The number of disintegrations per year of \(15.3\,\text{mg}\) sample of \({}^{60}_{27}\mathrm{Co}\) is \[ (\text{Mean life of }{}^{60}_{27}\mathrm{Co}\text{ is }7.65\text{ years and Avogadro's number }=6.024\times10^{23}\text{ mol}^{-1}) \]

• A. \(3624\times10^{16}\)
• B. \(798\times10^{15}\)
• C. \(2008\times10^{16}\)
• D. \(1258\times10^{16}\)

Hint:

For radioactive decay, \[ \boxed{ A=\lambda N=\frac{N}{\tau}, } \] where \(\tau\) is the mean life.

Answer 1

The Correct Option is C

Step 1: Calculate the number of nuclei. The number of nuclei is \[ N=\frac{m}{M}N_A. \] Here, \[ m=15.3\times10^{-3}\,\text{g}, \] \[ M=60\,\text{g mol}^{-1}, \] and \[ N_A=6.024\times10^{23}. \] Hence, \[ N=\frac{15.3\times10^{-3}}{60}\times6.024\times10^{23} =1.53612\times10^{20}. \]

Step 2: Use the activity formula. The number of disintegrations per year is \[ A=\lambda N=\frac{N}{\tau}, \] where \[ \tau=7.65\text{ years}. \] Therefore, \[ A=\frac{1.53612\times10^{20}}{7.65} =2.008\times10^{19}. \] Thus, \[ A=2008\times10^{16}. \] Hence, \[ \boxed{2008\times10^{16}.} \] Therefore, the correct option is \(\boxed{(C)}\).