Modern Physics Ques 55
- A hydrogen atom in its ground state is irradiated by light of wavelength $970 $ $\AA$. Taking $h c / e=1.237 \times 10^{-6}$ $ eVm$ and the ground state energy of hydrogen atom as $-13.6 $ $eV$, the number of lines present in the emission spectrum is
(2016 Adv.)
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Answer:
Correct Answer: 55.$(6)$
Solution:
Formula:
- Energy of incident light (in $eV$ )
$ E=\frac{12375}{970}=12.7 eV $
After excitation, let the electron jumps to $n$th state, then
$ \frac{-13.6}{n^{2}}=-13.6+12.7 $
Solving this equation, we get
$ n=4 $
$\therefore$ Total number of lines in emission spectrum,
$ =\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6 $
BETA
