A) \[n=1\to n=4\]
B) \[n=2\to n=4\]
C) \[n=2\to n=5\]
D) \[n=2\to n=3\]
Correct Answer: B
Solution :
| [b] Energy released for transition n = 2 to n = 1 of hydrogen atom |
| \[E=13.6{{Z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] |
| \[Z=1,{{n}_{1}}=1,{{n}_{2}}=2\] |
| \[E=13.6\times 1\times \left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] |
| \[E=13.6\times \frac{3}{4}eV\] |
| For \[H{{e}^{+}}\] ion z = 2 |
| [a] n = 1 to n = 4 |
| \[E=13.6\times {{2}^{2}}\times \left( \frac{1}{{{1}^{2}}}-\frac{1}{{{4}^{2}}} \right)=13.6\times \frac{15}{4}eV\] |
| [b] n = 2 to n = 4 |
| \[E=13.6\times {{2}^{2}}\times \left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=13.6\times \frac{3}{4}eV\] |
| [c] n = 2 to n = 5 |
| \[E=13.6\times {{2}^{2}}\times \left( \frac{1}{{{2}^{2}}}-\frac{1}{{{5}^{2}}} \right)=13.6\times \frac{21}{25}eV\] |
| [d] n = 2 to n = 3 |
| \[E=13.6\times {{2}^{2}}\times \left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=13.6\times \frac{5}{9}eV\] |
| Energy required for transition of \[H{{e}^{+}}\] for n = 2 to n = 4 matches exactly with energy released in transition of H for n = 2 to n = 1. |
You need to login to perform this action.
You will be redirected in
3 sec
