Question

A radioactive element has half-life period of $600yr$. After $3000yr$, what amount will remain?

• A. $\frac{1}{2}$
• B. $\frac{1}{16}$
• C. $\frac{1}{8}$
• D. $\frac{1}{32}$

Answer 1

The correct option is D $\frac{1}{32}$

We know that
$n=\frac{t}{T}$
Given, $t=3000yr,T=6000yr$
$n=\frac{3000}{6000}=5$
Then, $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$
$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^5=\frac{1}{32}$.