The ratio of the difference in energy between the first and the...

Energy of the (\mathrm{n}^{\text{th }}) Bohr's orbit is given by (E_n=-13.6 \times \frac{z^2}{n^2} e V) For H -atom, (\mathrm{Z}=1) The difference in energy between the first and the second Bohr orbit: $$\begin{aligned} & E_2-E_1=13.6 \times 1^2 \times\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\ & =13.6 \times \frac{3}{4} e V \end{aligned}$$) The $difference in energy between the second and the third Bohr orbit:$$\begin{aligned} & E_3-E_2=13.6 \times 1^2 \times\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\\ & =13.6 \times \frac{5}{36} e V \end{aligned}$$) Taking$ the $ratio, we get : (\frac{E_2-E_1}{E_3-E_2}=\frac{13.6 \times \frac{3}{4}}{13.6 \times \frac{5}{36}}=\frac{27}{5})