Question

One mole of radium has an activity of 1/3.7 kilo curie. Its decay 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

Answer: (A)

$\frac{1}{6}\times {10}^{-10}{s}^{-1}$

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

The mole of radium is $1$.

The decay constant is given as,

$\lambda =\frac{A}{N}$

$=\frac{\left(\frac{1}{3.7\times {10}^{-3}}\times 3.7\times {10}^{10}\right)}{6.023\times {10}^{23}}$

$=\frac{1}{6.023}\times {10}^{-10}{s}^{-1}$

Thus, the decay constant is $\frac{1}{6.023}\times {10}^{-10}{s}^{-1}$.