A) \[3.2\times {{10}^{15}}z\]
B) \[32\times {{10}^{15}}Hz\]
C) \[1.6\times 15\text{ }Hz\]
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 level\[{{n}_{2}}\]to \[{{n}_{1}}\]is \[v=Rc{{Z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] where,\[{{n}_{1}}\]is lower level,\[{{n}_{2}}\]is higher level, R is Rydberg constant; c is velocity of light and Z is atomic number of atom. When transition takes place from \[{{n}_{2}}=2\]to\[{{n}_{1}}=1\],then \[2.7\times {{10}^{15}}=Rc{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] ?(i) When transition takes place from\[{{n}_{2}}=3\]to \[{{n}_{1}}=1,\]frequency be \[v\] \[v=Rc{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] ?(ii) From Eqs. (i) and (ii), we get \[v=\frac{32\times 2.7\times {{10}^{15}}}{27}\] \[v=3.2\times {{10}^{15}}Hz\]
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