Chemistry - Radioactivity Question with Solution | TestHub

(C), (D)

According to given condition

${\mathrm{T}}_{1/2}={\mathrm{T}}_{\mathrm{av}}$

$\frac{0.693}{{\lambda }_{\mathrm{X}}}=\frac{1}{{\lambda }_{\mathrm{Y}}},{\lambda }_{\mathrm{Y}}>{\lambda }_{\mathrm{X}}$

Also $-\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$

i.e, $\frac{\mathrm{dN}}{\mathrm{dt}}\propto \lambda$ (it is given that N is same initially)

So Y will decay faster than X initially

Rate of decay for X and Y are given as

${\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}_{\mathrm{X}}={\lambda }_{\mathrm{X}}{\mathrm{N}}_{\mathrm{X}}$

${\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}_{\mathrm{Y}}={\lambda }_{\mathrm{Y}}{\mathrm{N}}_{\mathrm{Y}}$ and ${\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}_{\mathrm{X}}={\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}_{\mathrm{Y}}$

$\therefore {\lambda }_X{\mathrm{N}}_X={\lambda }_Y{\mathrm{N}}_Y$ and $\frac{{\mathrm{N}}_X}{{\mathrm{N}}_Y}=\frac{{\lambda }_Y}{{\lambda }_Y}$