A) 1.5eV
B) 0.85 eV
C) 3.4 eV
D) 1.9eV
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, \[\Delta E={{E}_{3}}-{{E}_{2}}=-\frac{13.6}{9}-\left( -\frac{13.6}{4} \right)\] \[=1.9eV\] (approximately)You need to login to perform this action.
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