A) \[{{n}_{2}}=4,{{n}_{1}}=3\]
B) \[{{n}_{2}}=5,{{n}_{1}}=3\]
C) \[{{n}_{2}}=4,{{n}_{1}}=2\]
D) \[{{n}_{2}}=4,{{n}_{1}}=1\]
Correct Answer: C
Solution :
As six different wavelengths are emitted from state \[{{n}_{2}},\]we have \[\frac{{{n}_{2}}({{n}_{2}}-1)}{2}=6\Rightarrow {{n}_{2}}=4\] Since, the energy of some emitted photons is less that of absorbed photon, \[{{n}_{1}}\ne 3.\]Since, the energy of some emitted photons is more than that of absorbed photon, \[{{n}_{1}}\ne 1.\] \[\therefore \] \[{{n}_{1}}=2\] Hence, the correction option is (c).You need to login to perform this action.
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