Modern Physics 1 Question 45

42. An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number nf.vi and vf are respectively the initial and final potential energies of the electron. If vivf=6.25, then the smallest possible nf is

(2017 Adv.)

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Solution:

  1. Potential energy of hydrogen atom (Z=1) in nth orbit (in eV )

$$ \begin{aligned} PE & =-\frac{27.2}{n^{2}} \ \frac{v _f}{v _i} & =\frac{-\frac{27.2}{n _f^{2}}}{-\frac{27.2}{n _i^{2}}}=\frac{1}{6.25} \



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