The ratio of transition energy of an electron from 2nd to 1st Bohr's orbit to that of an electron from 4th to 3rd Bohr's orbit in the hydrogen atom is approximately equal to -

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Solution

The correct option is B 15
The energy of electron in $n^{t h}$ orbit for hydrogen atom is given by:
$E_{n} = \frac{- 2 π^{2} m e^{4}}{n^{2} h^{2}}$
Therefore $E_{2} \to E_{1}$ = $\frac{- 2 π^{2} m e^{4}}{2^{2} h^{2}}$ –
$\frac{- 2 π^{2} m e^{4}}{1^{2} h^{2}}$
= $\frac{- 2 π^{2} m e^{4}}{h^{2}} \times (\frac{1}{4} - \frac{1}{1})$

For $E_{4} \to E_{3}$ = $\frac{- 2 π^{2} m e^{4}}{4^{2} h^{2}}$ –
$\frac{- 2 π^{2} m e^{4}}{3^{2} h^{2}}$
= $\frac{- 2 π^{2} m e^{4}}{h^{2}} \times (\frac{1}{16} - \frac{1}{9})$

Therefore, $\frac{E_{2} \to E_{1}}{E_{4} \to E_{3}}$ = $\frac{\frac{- 2 π^{2} m e^{4}}{h^{2}} \times (\frac{1}{4} - \frac{1}{1})}{\frac{- 2 π^{2} m e^{4}}{h^{2}} \times (\frac{1}{16} - \frac{1}{9})}$
= $\frac{\frac{1}{4} - \frac{1}{1}}{\frac{1}{16} - \frac{1}{9}}$

= $\frac{\frac{- 3}{4}}{\frac{- 7}{144}}$

= $\frac{144 \times 3}{28}$

=15.4

$\approx$ 15

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