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(B) The energy of the $n^{th}$ Bohr orbit is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For the hydrogen atom $(Z=1)$, the energy differ

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$27/5$

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(B) The energy of the $n^{th}$ Bohr orbit is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For the hydrogen atom $(Z=1)$, the energy difference between orbits $n_1$ and $n_2$ is $\Delta E = 13.6 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2}) 	ext{ eV}$.Difference in energy between the first $(n=1)$ and second $(n=2)$ orbits:$\Delta E_{1-2} = 13.6 	imes (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 	imes (1 - \frac{1}{4}) = 13.6 	imes \frac{3}{4} 	ext{ eV}$.Difference in energy between the second $(n=2)$ and third $(n=3)$ orbits:$\Delta E_{2-3} = 13.6 	imes (\frac{1}{2^2} - \frac{1}{3^2}) = 13.6 	imes (\frac{1}{4} - \frac{1}{9}) = 13.6 	imes \frac{5}{36} 	ext{ eV}$.The required ratio is $\frac{\Delta E_{1-2}}{\Delta E_{2-3}} = \frac{13.6 	imes (3/4)}{13.6 	imes (5/36)} = \frac{3}{4} 	imes \frac{36}{5} = \frac{27}{5}$.

The ratio of the difference in energy between the first and second Bohr orbits to that between the second and third orbits is

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