A) \[~n=1\]
B) \[~n=2\]
C) \[~n=3\]
D) \[~n=4\]
Correct Answer: C
Solution :
\[{{E}_{n}}=\frac{c{{z}^{2}}}{{{n}^{2}}}.\] For \[n=1\]state of hydrogen, we have \[{{E}_{1}}=c\]and for the nth state of \[L{{i}^{++}},\]we have \[{{E}_{n}}=\frac{c{{.3}^{2}}}{{{n}^{2}}}=\frac{9c}{{{n}^{2}}}\] For \[{{E}_{1}}={{E}_{n}}\]we require Hence, the correction option is [c]. \[C=\frac{9C}{{{n}^{2}}}\]which gives \[n=3.\]Hence \[n=3.\]
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