A) \[3.2\times {{10}^{15}}Hz\]
B) \[32\times {{10}^{15}}Hz\]
C) \[1.6\times {{10}^{15}}Hz\]
D) \[16\times {{10}^{15}}Hz\]
Correct Answer: A
Solution :
The frequency v of the emitted electromagnetic radiation, when a hydrogen atom de-excites from the level \[{{n}_{2}}\] to \[{{n}_{1}}\] is \[v=RC{{Z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] When transition takes place from \[{{n}_{2}}=2\] to\[{{n}_{1}}=1\], then \[2.7\times {{10}^{15}}=RC{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] When transition takes place from \[{{n}_{2}}=3\] to\[{{n}_{1}}=1\], then let the frequency be v. \[\therefore \] \[v=RC{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] From Eqs. (i) and (ii), we get \[v=\frac{32\times 2.7\times {{10}^{15}}}{27}=3.2\times {{10}^{15}}Hz\]
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