A) \[\frac{243}{32}f\]
B) \[\frac{141}{32}f\]
C) \[\frac{81}{32}f\]
D) \[\frac{63}{32}f\]
Correct Answer: A
Solution :
[a] \[E=hf=13.6{{Z}^{2}}\left( \frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}} \right)\] \[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{{{3}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)}{{{1}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right)}\Rightarrow {{f}_{2}}=\frac{243}{32}f.\]You need to login to perform this action.
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