• question_answer The energy of electron in the first energy level is $-21.79\times {{10}^{-12}}\,\text{erg}$ per atom. The energy of electron in second energy level is A) $-15.47\times {{10}^{-12}}\,\text{erg}\,\text{ato}{{\text{m}}^{-1}}$B) $-0.00547\times {{10}^{-12}}\,\text{erg}\,\text{ato}{{\text{m}}^{-1}}$C) $-4.557\times {{10}^{-12}}\,\text{erg}\,\text{ato}{{\text{m}}^{-1}}$D) $-5.447\times {{10}^{-12}}\text{erg}\,\text{ato}{{\text{m}}^{-1}}$

$(E)=-\frac{{{Z}^{2}}}{{{n}^{2}}}\times 2.179\times {{10}^{-11}}\,\text{erg/atom}\text{.}$ For first energy level, $n=1.$ ${{E}_{1}}=-\frac{{{Z}^{2}}}{{{1}^{2}}}\times 2.179\times {{10}^{-11}}\,\text{erg/atom}\text{.}$ For secondary energy level, $n=2.$             ${{E}_{2}}=-\frac{{{Z}^{2}}}{{{2}^{2}}}\times 2.179\times {{10}^{-11}}\,\text{erg/atom}\text{.}$             On dividing ${{E}_{2}}$by ${{E}_{1}},$we obtain             $\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{-{{Z}^{2}}\times 2.179\times {{10}^{-11}}/{{2}^{2}}}{-{{Z}^{2}}\times 2.179\times {{10}^{-11}}/{{1}^{2}}}=\frac{1}{4}$ ${{E}_{2}}=\frac{{{E}_{1}}}{4}=\frac{-21.79\times {{10}^{-12}}}{4}\text{erg/atom}\text{.}$ ${{E}_{2}}=-5.447\times {{10}^{-12}}\,\text{erg/atom}\text{.}$ Hence, the correct option is [d].