Question

The radioactive sources $A$ and $B$ of half lives of $8\text{hr}$ and $6\text{hr}$ respectively, initially contain the same number of radioactive atoms. At the end of $2\text{hours}$, their rates of disintegration are in the ratio

• A. $4:1$
• B. $2:1$
• C. $\sqrt{2}:1$
• D. $1:1$

Answer 1

The correct option is A $4:1$

Given:
Half-life of radioactive element
$T_A=8hr$
$T_B=6hr$

Time period t = 2 days (48 hours) As we know that radio activity is given by

$N=\frac{N_0}{2^n}$

Here, $n=\frac{t}{T}$

As given that $N_0$ is same,

$\therefore N_A=\frac{N_0}{2^{\left(48/8\right)}}=\frac{N_0}{2^6}$

$N_B=\frac{N_0}{2^{\left(48/6\right)}}=\frac{N_0}{2^8}$

Now ratio is $\frac{N_A}{N_B}=\frac{2^8}{2^6}=\frac{4}{1}$