A) 1.5 eV
B) 0.85 eV
C) 3.4 eV
D) 1.9 eV
Correct Answer: D
Solution :
Given: \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] |
Energy of photon ejected when electron jumps |
from\[n=3\] state to \[n=2\] state is given by |
\[\Delta E={{E}_{3}}-{{E}_{2}}\] |
\[\therefore \] \[{{E}_{3}}=-\frac{13.6}{{{(3)}^{2}}}eV=-\frac{13.6}{9}eV\] |
\[{{E}_{2}}=-\frac{13.6}{{{(2)}^{2}}}eV=-\frac{13.6}{4}eV\] |
So, \[\Delta E={{E}_{3}}-{{E}_{2}}=-\frac{13.6}{9}-\left( -\frac{13.6}{4} \right)\] |
\[=1.9\,eV\] (approximately) |
You need to login to perform this action.
You will be redirected in
3 sec