Chemistry - Atomic Structure Question with Solution | TestHub

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}\mathrm{m}{\mathrm{v}}^2$...(i)

$\text{mvr}=\frac{\mathrm{n}\mathrm{h}}{2\mathrm{\pi }}$

$\mathrm{v}=\frac{\mathrm{n}\mathrm{h}}{2\mathrm{\pi }\mathrm{m}\mathrm{r}}$...(ii)



Putting (ii) and (i)

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}\mathrm{m}\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\mathrm{\pi }}^2{\mathrm{m}}^2{\mathrm{r}}^2}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{8{\mathrm{\pi }}^2{\mathrm{r}}^2\mathrm{m}}$

Now$\mathrm{r}=\text{4}{\text{a}}_0$(since n$=2$) and${\mathrm{a}}_0=$Bohr's radius

$\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{8{\mathrm{\pi }}^2\times 16{\mathrm{a}}_0^2\mathrm{m}}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{128{\mathrm{\pi }}^2{\mathrm{a}}_0^2\mathrm{m}}$

Now$\mathrm{n}=2$

$\mathrm{K}.\mathrm{E}.=\frac{4{\mathrm{h}}^2}{128{\mathrm{\pi }}^2\mathrm{m}{\mathrm{a}}_0^2}=\frac{{\mathrm{h}}^2}{32{\mathrm{\pi }}^2\mathrm{m}{\mathrm{a}}_0^2}$