Question

The masses of two radioactive substances are same and their half lives are $1$ year and $2$ years respectively. The ratio of their activities after $6$ years will be

• A. $1:4$
  
• B. $1:6$
  
• C. $1:2$
  
• D. $1:3$
  

Answer 1

Answer: (A)

Given : $T_A=1$ years and $T_B=2$ years

Time of observation $t=6$ years

Let the initial mass (or nuclei) of two radioactive substance be $N_o$.

Initial activity of substance A $R_{o_A}={\lambda }_A{N}_o=\frac{\left(\ln 2\right)N_o}{T_A}$

Activity of substance A after time $t$, $R_A=R_{o_A}{e}^{\frac{-\left(\ln 2\right)t}{T_A}}$

Initial activity of substance B $R_{o_B}={\lambda }_B{N}_o=\frac{\left(\ln 2\right)N_o}{T_B}$

Activity of substance B after time $t$, $R_B=R_{o_B}{e}^{\frac{-\left(\ln 2\right)t}{T_B}}$

Thus ratio of activities after 6 years, we get $\frac{R_A}{R_B}=\frac{R_{o_A}}{R_{o_B}}{e}^{-\left(\ln 2\right)[\frac{t}{T_A}-\frac{t}{T_B}]}$

Or $\frac{R_A}{R_B}=\frac{T_B}{T_A}{e}^{-\left(\ln 2\right)[\frac{6}{1}-\frac{6}{2}]}$

Or $\frac{R_A}{R_B}=\frac{2}{1}{e}^{-3\ln 2}=2\times e^{\ln \frac{1}{8}}$

Or $\frac{R_A}{R_B}=2\times \frac{1}{8}$

$⟹$ $R_A:R_B=1:4$