Question

The count rate from a radioactive sample fails from 4.0 $\times {10}^6$ per second to 1.0 $\times {10}^6$ per second in 20 hours. What will be the count rate 100 hours after the beginning ?

Answer 1

Given $A_0=4\times {10}^6$ disintegrations /sec

A' = $1\times {10}^6$ disintegrations/sec

t = 20 hrs

$A^{\prime }=\frac{{\text{A}}_0}{2^{\frac{t}{\frac{t1}{2}}}}\Rightarrow 2^{\frac{t}{t\frac{1}{2}}}=\frac{A_0}{A^{\prime }}$

$\Rightarrow 2^{\frac{t}{t\frac{1}{2}}}=4$

$\Rightarrow \frac{t}{t\frac{1}{2}}=2$

$t\frac{1}{2}=2$ $t\frac{1}{2}=\frac{t}{2}=20\text{hrs.}/2=10\text{hrs.}$

$A^{\prime \prime }=\frac{A_0}{2^{\frac{t}{\frac{t1}{2}}}}$

$A^{\prime \prime }=\frac{4\times {10}^8}{2^{\frac{100}{10}}}$

$=0.00330625\times {10}^4$

= $3.9\times {10}^8\text{disintegrations/sec.}$