Chemistry - Atomic Structure Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{E}}_{\mathrm{n}}\left({\mathrm{Li}}^{2+}\right)=-19.62\times 10^{-18}\mathrm{J};\\ & {\mathrm{r}}_{\mathrm{n}}\left({\mathrm{Li}}^{2+}\right)=17.6\mathrm{pm}\\ & {\mathrm{E}}_{\mathrm{n}}\left({\mathrm{He}}^{+}\right)=-8.72\times 10^{-18}\mathrm{J};\\ & {\mathrm{r}}_{\mathrm{n}}\left({\mathrm{He}}^{+}\right)=26.4\mathrm{pm}\end{array}$

$\begin{array}{rl}& {\mathrm{E}}_{\mathrm{n}}\left({\mathrm{Li}}^{2+}\right)=-8.72\times 10^{-18}\mathrm{J}\\ & {\mathrm{r}}_{\mathrm{n}}\left({\mathrm{Li}}^{2+}\right)=26.4\mathrm{pm}\\ & {\mathrm{E}}_{\mathrm{n}}\left({\mathrm{He}}^{+}\right)=-19.62\times 10^{-18}\mathrm{J};\\ & {\mathrm{r}}_{\mathrm{n}}\left({\mathrm{He}}^{+}\right)=17.6\mathrm{pm}\end{array}$

$\begin{array}{rl}& {\mathrm{E}}_{\mathrm{n}}\left({\mathrm{Li}}^{2+}\right)=-19.62\times 10^{-16}\mathrm{J}\\ & {\mathrm{r}}_{\mathrm{n}}\left({\mathrm{Li}}^{2+}\right)=17.6\mathrm{pm}\\ & {\mathrm{E}}_{\mathrm{n}}\left({\mathrm{He}}^{+}\right)=-8.72\times 10^{-16}\mathrm{J}\\ & {\mathrm{r}}_{\mathrm{n}}\left({\mathrm{He}}^{+}\right)=26.4\mathrm{pm}\end{array}$

$\begin{array}{rl}& {\mathrm{E}}_{\mathrm{n}}\left({\mathrm{Li}}^{2+}\right)=-8.72\times 10^{-16}\mathrm{J}\\ & {\mathrm{r}}_{\mathrm{n}}\left({\mathrm{Li}}^{2+}\right)=17.6\mathrm{pm}\\ & {\mathrm{E}}_{\mathrm{n}}\left({\mathrm{He}}^{+}\right)=-19.62\times 10^{-16}\mathrm{J};\\ & {\mathrm{r}}_{\mathrm{n}}\left({\mathrm{He}}^{+}\right)=17.6\mathrm{pm}\end{array}$