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 :
\(\frac{E_2-E_1}{E_3-E_2}=\frac{13.6 \times \frac{3}{4}}{13.6 \times \frac{5}{36}}=\frac{27}{5}\)