A) \[{{E}_{n}}\] for \[{{A}^{(+Z-)}}={{Z}^{2}}\times {{E}_{n}}\] for H
B) \[{{E}_{n}}\] for \[{{A}^{(+Z-1)}}=Z\times {{E}_{n}}\] for H
C) \[{{E}_{n}}\] for \[{{A}^{(+Z-1)}}=\frac{1}{{{Z}^{2}}}\times {{E}_{n}}\] for H
D) \[{{E}_{n}}\] for \[{{A}^{(+Z-1)}}=\frac{1}{Z}\times {{E}_{n}}\] for H
Correct Answer: A
Solution :
\[{{E}_{He+}}={{E}_{H}}\times {{2}^{2}};{{E}_{Li}}^{2+}={{E}_{H}}\times {{3}^{2}}.......\]up to Z
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