Question

At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio $\frac{R_B}{R_A}$ of their activities after time t itself decays with time t as $e^{-3t}.$ If the half-life of A is $ln2$, the half-life of B is:

• A. $\frac{ln2}{2}$
• B. $2ln2$
• C. $\frac{ln2}{4}$
• D. $4ln2$

Answer 1

The correct option is C $\frac{ln2}{4}$

Half life of $A=ln2$
$t_{1/2}=\frac{\ell n2}{\lambda }$
${\lambda }_A=1$
at $t=0R_A=R_B$
$N_A{e}^{-\lambda AT}=N_B{e}^{-\lambda BT}$
$N_A=N_B$ at t = 0
at t= t $\frac{R_B}{R_A}=\frac{N_0{e}^{-{\lambda }_B^t}}{N_0{e}^{-{\lambda }_A^t}}$
$e^{-({\lambda }_B-{\lambda }_A)t}=e^{-3t}$
${\lambda }_B-{\lambda }_A=3$
${\lambda }_B=3+{\lambda }_A=4$
$t_{1/2}=\frac{\ell n2}{{\lambda }_B}=\frac{\ell n2}{4}$