A) 1 : 1
B) 1 : 3
C) 1 : 9
D) 1 : 16
Correct Answer: D
Solution :
| \[{{E}_{n}}={{E}_{0}}\frac{{{Z}^{2}}}{{{n}^{2}}}\] |
| \[\frac{{{E}_{H}}}{{{E}_{B{{e}^{3+}}}}}=\frac{1}{{{(1)}^{2}}}\times \frac{1}{{{(4)}^{2}}}=\frac{1}{16}\] |
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