Question

A radioactive substance of half life $6min$ is placed near a Geiger counter which is found to register $1024$ particles per minute. How many particles per minute will be register after $42min$?

• A. $4/min$
• B. $8/min$
• C. $5/min$
• D. $7/min$

Answer 1

The correct option is B $8/min$

Given : Half life of radioactive substance, $T_{1/2}=6$ $min$

Initial count rate, $R_o=1024$ /$min$

Let the count rate after 42 min be $R_f$.

Using: $\lambda t=ln\frac{R_o}{R_f}$ where, $\lambda =\frac{0.693}{T_{1/2}}$

$⟹$ $\frac{0.693}{T_{1/2}}t=ln\frac{R_o}{R_f}$

$⟹$ $\frac{0.693}{6}\times 42=ln\frac{1024}{R_f}$

$⟹$ $ln\frac{1024}{R_f}=4.851$

$⟹\frac{1024}{R_f}=e^{4.851}=127.87$

$⟹$ $R_f=\frac{1024}{127.87}=8$

Thus, final count rate is $8$ /$min$.