Question

Half-life period of a radioactive nucleus is $3$ hours. In $9$ hours its activity will be reduced by:

• A. $\frac{1}{9}$
• B. $\frac{1}{8}$
• C. $\frac{7}{8}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

$\frac{1}{8}$

Given that:-

Half-life period $\left(t_{1/2}\right)=3\text{hours}$

Total time taken $\left(t\right)=9\text{hours}$

As we know that,

$\frac{N}{N^0}=\frac{1}{2^n}$

where,

$N^0=$ Original no. of nuclides

$N=$ Reduced no. of nuclides

$n=\frac{t}{t_{1/2}}=\frac{9}{3}=3$

$\therefore N=\left(\frac{1}{2^3}\right)N^0$

$\Rightarrow N=\left(\frac{1}{8}\right)N^0$

Hence, the activity will be reduced by $\frac{1}{8}$ times.