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 $=\frac{100\times 6.023\times {10}^6}{6.023\times {10}^{23}}={10}^{-15}g$