$\frac{\Delta p_a}{\Delta p_e}=\frac{1}{2}$
The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is $\frac{1}{4}$
$m=2$
$\lambda_e=418$ nm

Solution:

\;\; \frac{1}{\lambda_a}=R\left(1-\;\; \frac{1}{16}\right)

\;\; \frac{1}{\lambda_e}=R\left(\;\; \frac{1}{m^2}-\;\; \frac{1}{4^2}\right)

\;\; \frac{\lambda_a}{\lambda_e} = \;\; \frac{ \left(\;\; \frac{1}{m^2} - \;\; \frac{1}{4^2} \right) }{ \;\; \frac{15}{16} } = \frac{1}{5}

\;\; \frac{1}{m^2} = \;\; \frac{1}{4^2} + \;\; \frac{3}{16}

\;\; \frac{1}{m^2} = \;\; \frac{1}{4} \Rightarrow m=2

⇒Kinetic energy

\alpha\;\; \frac{1}{n^2}

\;\; \frac{K_m}{K_1} = \;\; \frac{1}{2^2} \times 1 = \;\; \frac{1}{4}

13.6\left(\;\; \frac{1}{4} - \;\; \frac{1}{16}\right) = \;\; \frac{1242}{\lambda_e}

13.6\left(\;\; \frac{3}{16}\right) = \;\; \frac{1242}{\lambda_e}

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