Question

The graph represents the decay of a newlyprepared sample of radioactive nuclide X to a stable nuclide Y. The half-life of X is t. The growth curve for Y intersects the decay curve for X after time T. What is the time T?

• A. $ln(t/2)$
• B. $ln(2t)$
• C. $t/2$
• D. $t$

Answer 1

Answer: (D)

According to radioactive decay law, number of active nuclei of unstable nuclei at time $T$ is given by
$N_1=N_0{e}^{-\lambda T}....\left(1\right)$
and that for stable nuclei is given by
$N_2=N_0\left(1-e^{-\lambda T}\right)....\left(2\right)$
where symbols have their usual meanings. We have assumed $\lambda$ as same for both nuclei.
$\therefore$ When graph intersects then
$N_1=N_2$
$\Rightarrow N_0{e}^{-\lambda T}=N_0\left(1-e^{-\lambda T}\right)$
$\Rightarrow 2e^{-\lambda T}=1$
$\Rightarrow e^{\lambda T}=2$
$\Rightarrow \lambda T=ln2$
$\Rightarrow T=\frac{ln2}{\lambda }=t_{1/2}=$ half life
Given, $t_{1/2}=t$
$\therefore T=t.$