An electron from various excited states of hydrogen atom emit radiation to come to the ground state ${\lambda }_n, {\lambda }_g$ be the de Broglie wavelength of the electron the $n^{\text{th }}$ state and the ground state respectively. Let ${\wedge }_n$  be the wavelength of the emitted photon in the transition from the $n^{\text{th }}$ state to the ground state. For ( $A, B$ are constants

Momentum of electron in different states

        $p_n=\frac{h}{{\lambda }_n}, p_g=\frac{h}{{\lambda }_g}$

Kinetic energy, $K=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2}$

Total energy in an orbit of hydrogen atom,

$$\begin{gathered} E=-K=-\frac{h^2}{2m{\lambda }^2} \\ {E}_n-E_g=\frac{h^2}{2m}\left(\frac{1}{{\lambda }_g^2}-\frac{1}{{\lambda }_n^2}\right) \end{gathered}$$

$$\begin{gathered} \frac{h^2}{2m}\left(\frac{{\lambda }_m^2-{\lambda }_g^2}{{\lambda }_g^2{\lambda }_n^2}\right)=\frac{hc}{{\Lambda }_n} \\ \Rightarrow {\Lambda }_n=\frac{2mc}{h}\left(\frac{{\lambda }_g^2{\lambda }_n^2}{{\lambda }_m^2-{\lambda }_g^2}\right) \\ \Rightarrow {\Lambda }_n=\frac{2mc{\lambda }_g^2}{h}{\left[1-\frac{{\lambda }_g^2}{{\lambda }_n^2}\right]}^{-1}=\frac{2mc{\lambda }_g^2}{h}\left[1+\frac{{\lambda }_g^2}{{\lambda }_m^2}\right] \left\{\because {\lambda }_g\ll {\lambda }_n\right\} \\ \Rightarrow {\Lambda }_n=\frac{2mc{\lambda }_g^2}{h}+\left(\frac{2mc{\lambda }_g^4}{h}\right)\frac{1}{{\lambda }_m^2}=A+\frac{B}{{\lambda }_m^2} \end{gathered}$$

$\text{ where }A\text{ and }B\text{ are }A=\frac{2mc{\lambda }_g^2}{h},B=\frac{2mc{\lambda }_g^4}{h}$

Hence, the correct answer is (A).