Question

Activities of three radioactive substances $A$, $B$ and $C$ are represented by the curves $A$, $B$ and $C$ , in the figure. Then their half-lives $T_{1/2}\left(A\right):T_{1/2}\left(B\right):T_{1/2}\left(C\right)$ are in the ratio:

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

Answer 1

The correct option is D

$2:1:3$

Radioactive Decay:

The total decay rate $R$ of a sample of one or more radioactive nucleus is called the activity of that sample.

$R=-\frac{\partial N}{\partial t}=\lambda N{e}^{-\lambda t}=R_0{e}^{-\lambda t}$

$R=R_0{e}^{-\lambda t}$

where $N$ is amount of radioactive substance, $\lambda$ is the decay constant

Taking log both side,

$\ln R=\lambda t\ln R_0$

the decay constant is, $\lambda =\frac{\ln R}{t\ln R_0}$

for the A nuclear substance,

${\lambda }_A=\frac{6}{10\ln R_0}$

for the B nuclear substance,

${\lambda }_B=\frac{6}{5\ln R_0}$

for the B nuclear substance,

${\lambda }_C=\frac{2}{5\ln R_0}$

Also, half life is $T_{1/2}=\frac{\ln 2}{\lambda }$

So, $T_{1/2}\left(A\right):T_{1/2}\left(B\right):T_{1/2}\left(C\right)=\frac{1}{{\lambda }_A}:\frac{1}{{\lambda }_B}:\frac{1}{{\lambda }_C}$

$T_{1/2}\left(A\right):T_{1/2}\left(B\right):T_{1/2}\left(C\right)=\frac{10}{6}:\frac{5}{6}:\frac{5}{2}$

$T_{1/2}\left(A\right):T_{1/2}\left(B\right):T_{1/2}\left(C\right)=2:1:3$

Hence, option D is correct