Modern Physics 1 Question 47
43. 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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Solution:
- 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,
$$ =\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6 $$
