A) \[III\]
B) \[IV\]
C) \[I\]
D) \[II\]
Correct Answer: A
Solution :
\[E=Rhc\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[{{E}_{(4\to 3)}}=Rhc\left[ \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right]\] \[=Rhc\left[ \frac{7}{9\times 16} \right]=0.05\,\,Rhc\] \[{{E}_{(4\to 2)}}=Rhc\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]\] \[=Rhc\left[ \frac{3}{16} \right]=0.2\,\,Rhc\] \[{{E}_{(2\to 1)}}=Rhc\left[ \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right]\] \[=Rhc\left[ \frac{3}{4} \right]=0.75\,\,Rhc\] \[{{E}_{(1\to 3)}}=Rhc\left[ \frac{1}{{{(3)}^{2}}}-\frac{1}{{{(1)}^{2}}} \right]\] \[=-\frac{8}{9}Rhc=-0.9\,\,Rhc\] Thus, transition III gives most energy. Transition I represents the absorption of energy.
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