Question

The activity of a radioactive sample drops to 1/64th of its original value in 2 hr. The decay constant for the sample is :

• A. $5.775\times {10}^{-4}{s}^{-1}$
• B. $5.712\times {10}^{-4}{s}^{-1}$
• C. $5.722\times {10}^{-4}{s}^{-1}$
• D. none of these

Answer 1

Answer: (A)

$5.775\times {10}^{-4}{s}^{-1}$

$\frac{r_0}{r_1}=64$

$r_0\propto N_0$

$r_1\propto N_1$

$\frac{r_0}{r_1}=\frac{N_0}{N_1}=64$

$N_1=\frac{N_0}{64}=\frac{N_0}{2^6}$

$n=6$, i.e. the number of half-lives.

$Time=t_{1/2}\times n$

$(\therefore t=2hr)$

$2\times 60\times 60=t_{1/2}\times 6$

$t_{1/2}=1200s$

$K=\frac{0.693}{1200}=5.775\times {10}^{-4}{s}^{-1}$