A) n = 2 to n = 1
B) n = 3 to n = 2
C) n = 4 to n = 3
D) n = 3 to n = 1
Correct Answer: A
Solution :
\[hv=\Delta E=13.6{{z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[{{v}_{He+}}={{v}_{H}}\times {{z}^{2}}\left( \frac{1}{{{\left( \frac{{{n}_{1}}}{2} \right)}^{2}}}-\frac{1}{{{\left( \frac{{{n}_{2}}}{2} \right)}^{2}}} \right)\] \[={{v}_{H}}\left( \frac{1}{{{\left( \frac{2}{2} \right)}^{2}}}-\frac{1}{{{\left( \frac{4}{2} \right)}^{2}}} \right)\] For H-atom\[{{n}_{1}}=1,{{n}_{2}}=2\]
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