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:

  1. 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,

$$ =\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6 $$



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