A) \[{{n}_{1}}=2\to {{n}_{2}}=1\]
B) \[{{n}_{1}}=3\to {{n}_{2}}=1\]
C) \[{{n}_{1}}=10\to {{n}_{2}}=2\]
D) \[{{n}_{1}}=100\to {{n}_{2}}=3\]
Correct Answer: B
Solution :
\[{{E}_{1}}=-13.6\,eV,\,\,{{E}_{2}}=-3.4\,eV,\,\,{{E}_{3}}=-1.51\,eV,\] \[{{E}_{10}}=-0.136\,eV,\,\,{{E}_{100}}=-0.0013\,eV\] \[{{E}_{21}}=-13.6-(-3.4)=-10.2eV\] \[{{E}_{31}}=-13.6-(-1.51)=-12.09\,eV\] \[{{E}_{10.2}}=-3.4-(-0.136)=-3.264\,\,eV\] \[{{E}_{100,2}}=-1.51-(-0.0013)=-1.508\,\,eV\] Hence, \[{{E}_{31}}\] is the most energetic transition.
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