Atoms Ques 26
26. Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E=-13.6 / n^{2} eV$. The energy of photon ejected when the electron jumps from $n=3$ state to $n=2$ state of hydrogen is approximately
(a) $1.9 eV$.
(b) $1.5 eV$.
(c) $0.85 eV$
(d) $3.4 eV$
[2004]
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Answer:
Correct Answer: 26.(a)
Solution:
- (a) $\Delta E=E_3-E_2$
$=\frac{-13.6}{3^{2}}+\frac{13.6}{2^{2}}$
$=13.6{{\frac{1}{4}-\frac{1}{9}}} eV=1.9 eV$
BETA
