A) n = 1 to n = 2
B) n = 2 to n = 1
C) n = 2 to n = 6
D) n = 6 to n = 2
Correct Answer: B
Solution :
Energy level of H-atom are given by \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] Photons are emitted only when electron jump from higher energy level (higher n-value) to lower energy level (lower n-value). So alternative (a) and (c) are wrong. Energy difference from n = 2 to n - 1 level is \[\Delta {{E}_{2\to 1}}=13.6\,\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)eV\] \[=13.6\times \frac{3}{4}=10.2\,eV\] Energy difference from n = 6 to n = 2 level is \[\Delta {{E}_{6\to 2}}=13.6\,\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{6}^{2}}} \right)eV\] \[=13.6\times \left( \frac{1}{4}-\frac{1}{36} \right)eV\] \[=13.6\times \frac{2}{9}eV\] = 3.02 eV Thus, it is evident that difference is larger for n = 2 to n = 1 transition. Hence, maximum energy photon will be emitted during transition from n = 2 to n = 1.
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