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 and 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 |
| \[\Delta {{E}_{6\to 2}}=13.6\,\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{6}^{2}}} \right)eV\] |
| \[=13.6\times \left( \frac{1}{4}-\frac{1}{36} \right)eV\] |
| \[=13.6\times \frac{2}{9}eV\] |
| \[=3.02\text{ }eV\] |
| 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 \[n=1\]. |
You need to login to perform this action.
You will be redirected in
3 sec
