A) \[3\to 1\]
B) \[4\to 2\]
C) \[4\to 3\]
D) \[3\to 2\]
Correct Answer: A
Solution :
From Bohrs postulate, energy of electron in nth orbit is given by \[E=-\frac{M{{Z}^{2}}{{e}^{4}}}{8\varepsilon _{0}^{2}{{h}^{2}}}\left( \frac{1}{{{n}^{2}}} \right)\] When electron jumps from some higher energy state\[{{n}_{2}}\]to a lower energy state\[{{n}_{1}},\]the energy difference between these states is \[{{E}_{2}}-{{E}_{1}}\propto \left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] From Bohrs third postulate, the frequency v of the electromagnetic wave is \[v=\frac{{{E}_{2}}-{{E}_{1}}}{h}\propto \left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] First case \[{{n}_{1}}=1,{{n}_{2}}=3\] \[{{v}_{1}}\propto \left( 1-\frac{1}{9} \right)\propto \frac{8}{9}\] Second case \[{{n}_{1}}=2,{{n}_{2}}=4\] \[{{v}_{2}}\propto \left( \frac{1}{4}-\frac{1}{16} \right)\propto \frac{3}{16}\] Third case \[{{n}_{1}}=3,{{n}_{2}}=4\] \[\therefore \] \[{{v}_{3}}\propto \left( \frac{1}{9}-\frac{1}{16} \right)\propto \frac{7}{144}\] Fourth case \[{{n}_{1}}=2,{{n}_{2}}=3\] \[\therefore \] \[{{v}_{4}}\propto \left( \frac{1}{4}-\frac{1}{9} \right)\propto \frac{5}{36}\] \[{{v}_{1}}>{{v}_{2}}>{{v}_{4}}>{{v}_{3}}\] Hence, transition \[3\to 1\]has higher frequency.
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