Question

A radioactive nucleus decays by two different processes. The half-life for the first process is $10s$ and that for the second is $100s$. The effective half-life of the nucleus is close to:

• A. $55sec$
• B. $6sec$
• C. $12sec$
• D. $9sec$

Answer 1

The correct option is D

$9sec$

Step 1: Given data

Half life for first process,$T_1=10s$

Half life for the second process, $T_2=100s$

Let the effective half life of two processes $=T_{eq}$

Let the decay constant for the first process $={\lambda }_1$

Let the decay constant for the second process $={\lambda }_2$

Let the effective decay constant of two processes $={\lambda }_{eq}$

Step 2: Find effective half-life of the nucleus

The decay constant for the first process is

${\lambda }_1=[\ln 2/T_1]$

The decay constant for the second process is

${\lambda }_2=[\ln 2/T_2]$

The effective decay constant of two processes is

${\lambda }_{eq}=\ln 2/T_{eq}$

The effective decay constant of two processes can also be given by

${\lambda }_{eq}={\lambda }_1+{\lambda }_2$

Put all the values in above formulae, we get

$$\begin{gathered} (\ln 2/T_{eq})=(\ln 2/T_1)+(\ln 2/T_2) \\ (1/T_{eq})=(1/10)+(1/100) \\ (1/T_{eq})=(10+1)/100 \\ (1/T_{eq})=11/100 \\ {T}_{eq}=100/11 \\ {T}_{eq}=9s \end{gathered}$$

Hence, option D is correct.