Chemistry - Atomic Structure Question with Solution | TestHub

Energy of G.S. of$\mathrm{H}{\mathrm{e}}^{+}={\mathrm{E}}_{\mathrm{H}}\times {\mathrm{Z}}^2$

$=-13.6\times 4=-54.4\text{eV}$

Ionisation potential of$\mathrm{H}{\mathrm{e}}^{+}={\mathrm{E}}_{\mathrm{\infty }}-{\mathrm{E}}_1$

$=-\left(-54.4\right)=54.4\text{eV}$

T.E. of H of 1 orbit$=-13.6\text{eV}$

2 T.E. = P.E. and$\text{K}\text{.E}=-\text{ T}\text{. E}$

$\text{P}\text{.E}\text{.}=-2\times 13.6=-27.2\text{eV}$

T.E. of H of II excited state$=\frac{-13.6}{9}\times 4$

of$\text{He}=-6.04\text{eV}$

$\text{K}\text{.E}\text{.}=-\text{(T}\text{.E}\text{.)}=-\left(-6.04\text{eV}\right)=+6.04\text{eV}$