d
(D) \({E_n}\,\, = \,\, - \frac{{2{\pi ^2}{Z^2}m{e^4}}}{{{n^2}{h^2}}}\,\, = \,\, - K\frac{{{z^2}}}{{{n^2}}}\,\,\,\)જ્યાં,\(\,K\,\, = \,\,\frac{{2{\pi ^2}\,m{e^4}}}{{{n^2}}}\)

\(He^+\) ની આયાનીકરણ-ઉર્જા \(={{\text{E}}_\infty }{\text{   -  }}{{\text{E}}_{\text{1}}}{\text{  }} = \,\,0\,\, - \,\left( { - K\,\frac{{{{(2)}^2}}}{{{{(1)}^2}}}} \right)\,\, = \,\,4\,K\)

\(4K  =  19.6  \times 10^{- 18}\) જૂલ \(\therefore \,\,K\,\, = \,\,\frac{{19.6\, \times \,{{10}^{ - 18}}}}{4}\,\, =  4.9 \times 10^{- 18}\) જૂલ પરમાણુ 

\(Li^{2+}\) માટે \( Z = 3\) અને  \(n = 1 \)

\(\therefore \,\,{E_1}\, = \, - K\,\frac{{{Z^2}}}{{{n^2}}}\,\, = \,\, - \,\frac{{4.9\, \times \,{{10}^{18}}\, \times \,9}}{1}\, =-  4.41 \times 10^{ - 17}\) જૂલ પરમાણુ