Question

In a radioactive material, fraction of active material remaining after time $t$ is $9 / 16 .$ The fraction that was remaining after $t / 2$ is :

• A. $\frac{3}{4}$
• B. $\frac{7}{8}$
• C. $\frac{4}{5}$
• D. $\frac{3}{5}$

Answer 1

Answer: (A)

$\frac{3}{4}$

Answer 2

First order decay
$N ( t )= N _{0} e ^{-\lambda t }$
Given $N ( t ) / N _{0}=9 / 16= e ^{-\lambda t }$
Now, $N ( t / 2)= N _{0} e ^{-\lambda t / 2}$
$\frac{ N ( t / 2)}{ N _{0}}=\sqrt{ e ^{-\lambda t }}$
$=\sqrt{9 / 16}$
$N ( t/ 2)=3 / 4 N _{0}$