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) \[-5.447\times {{10}^{-12}}\text{erg}\,\text{ato}{{\text{m}}^{-1}}\]
Correct Answer: D
Solution :
\[(E)=-\frac{{{Z}^{2}}}{{{n}^{2}}}\times 2.179\times {{10}^{-11}}\,\text{erg/atom}\text{.}\] 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].
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