Question

The activity of ratio altitude $ ({{X}^{100}}) $ is 6.023 curie. If its disintegration constant is $ 3.7\times {{10}^{4}}{{\sec }^{-1}}, $ the mass of X is :

• A. $ {{10}^{-23}}g $
• B. $ {{10}^{-15}}g $
• C. $ {{10}^{-16}}g $
• D. $ {{10}^{-14}}g $

Answer 1

Answer: (B)

We know that 1 Curie $ 3.7\,\,\times \,\,{{10}^{10}} $ disintegration per sec So, activity $ =6.023\times 3.7\times {{10}^{10}}\,\text{dps} $ $ \lambda =3.7\times {{10}^{14}}\,{{\sec }^{-1}} $ $ N=\frac{Activity}{\lambda }=\frac{6.023\times 3.7\times {{10}^{-10}}}{3.7\times {{10}^{4}}} $ $ =6.023\times {{10}^{6}}\text{atoms} $ $ \because $ Weight of $ 6.023\times {{10}^{23}} $ atoms = 100 g $ \therefore $ Weight of $ 6.023\times {{10}^{6}} $ atoms will be $ =\frac{100\times 6.023\times {{10}^{6}}}{6.023\times {{10}^{23}}}={{10}^{-15}}\,g $