Question

One mole of radium has an activity of 1/3.7 killo curie. Its decay constant will be

• A. $\frac{1}{6}\times {10}^{-10}{s}^{-1}$
• B. ${10}^{-10}{s}^{-1}$
• C. ${10}^{-11}{s}^{-1}$
• D. ${10}^{-8}{s}^{-1}$

Answer 1

The correct option is C ${10}^{-11}{s}^{-1}$

Number of mole of radium, $n=1$

The number of nuclei of radium, $N=N_a=avagdaronumber=6.023\times {10}^{23}$

Activity, $A=\lambda N$

$\because 1curie=3.7\times {10}^{10}decay/\sec$

$\frac{1}{3.7}kilocurie=0.27kilocurie=9.99\times {10}^{12}dacay/\sec$

So, $\lambda =\frac{A}{N}=\frac{9.99\times {10}^{12}}{6.023\times {10}^{23}}=1.65\times {10}^{-11}$