A) \[400nm\]
B) \[421nm\]
C) \[434nm\]
D) \[43.4nm\]
Correct Answer: C
Solution :
\[{{n}_{i}}=5\]and \[{{n}_{f}}=2\] \[\Delta E=2.18\times {{10}^{-18}}\left( \frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}} \right)\] \[\Delta E=2.18\times {{10}^{-18}}\left( \frac{1}{{{5}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] \[=-4.58\times {{10}^{-19}}J\] \[v=\frac{\Delta E}{h}=\frac{4.58\times {{10}^{-19}}}{6.626\times {{10}^{-34}}}\] \[=6.91\times {{10}^{14}}Hz\] \[\lambda =\frac{c}{v}=\frac{3.0\times {{10}^{8}}m{{s}^{-1}}}{6.91\times {{10}^{14}}Hz}\] \[=434nm\]
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