A) \[a=\frac{9}{4}\]
B) \[b=\frac{27}{5}\]
C) \[c=\frac{5}{27}\]
D) \[c=\frac{1}{a}=\frac{5}{27}\]
Correct Answer: C
Solution :
First transition is from\[n=3\]to\[n=2,\]second transition is from\[n=2\]to\[n=1\] \[\therefore \] \[\frac{{{E}_{1}}}{{{E}_{2}}}=c=\frac{\frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}}}{\frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}}}\] or \[c=\frac{5/36}{3/4}=\frac{5}{36}\times \frac{4}{3}=\frac{5}{27}\] As, \[p=\frac{E}{c'}\], therefore \[\frac{{{p}_{1}}}{{{p}_{2}}}=b=\frac{{{E}_{1}}}{{{E}_{2}}}=c,\]i.e., \[b=c=\frac{5}{27}\] As, \[E=\frac{hc}{\lambda }\] \[\therefore \] \[\lambda \propto \frac{1}{E}\]
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