Question

A radioactive element has half-life period $1600yr$. After $6400yr$, what part of element will remain?

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{16}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

From Rutherford Soddy law, the number of atoms left after $n$ half-lives is
$N=N_0{\left(\frac{1}{2}\right)}^n$
where, $n=\frac{\text{ time }(t)}{{\text{ half-life(T }}_{1/2})}$
Given, $t=6400yr,T=1600yr$
$\therefore n=\frac{6400}{1600}=4$
So, $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n={\left(\frac{1}{2}\right)}^4=\frac{1}{16}$
Hence, $\frac{1}{16}$ part of element will remain.