Atoms - Result Question 30
32. 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
«««< HEAD
======= ####32. 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
c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (a) $1.9 eV$.
(b) $1.5 eV$.
(c) $0.85 eV$
(d) $3.4 eV$
[2004]
Show Answer
Answer:
Correct Answer: 32. (a)
Solution:
- (a) $\Delta E=E_3-E_2$
$=\frac{-13.6}{3^{2}}+\frac{13.6}{2^{2}}$
$=13.6{\frac{1}{4}-\frac{1}{9}} eV=1.9 eV$
