An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let $\lambda_n$, $\lambda_g$ be the de Broglie wavelength of the electron in the $n^{\text{th}}$ state and the ground state respectively. Let $\Lambda_n$ be the wavelength of the emitted photon in the transition from the $n^{\text{th}}$ state to the ground state. For large $n$ ($A$, $B$ are constants):
Answer: (B) $\Lambda_n \approx A + \dfrac{B}{\lambda_n^2}$
In Bohr's model $p\propto\dfrac1n$, so $\lambda_n = n\lambda_g$, i.e. $\dfrac1{n^2} = \dfrac{\lambda_g^2}{\lambda_n^2}$.
The photon wavelength:
$$\frac1{\Lambda_n} = R\left(1 - \frac1{n^2}\right) \;\Rightarrow\; \Lambda_n = \frac1R\left(1 - \frac1{n^2}\right)^{-1} \approx \frac1R\left(1 + \frac1{n^2}\right)$$
for large $n$. Substituting,
$$\Lambda_n \approx \frac1R + \frac{\lambda_g^2}{R}\cdot\frac1{\lambda_n^2} = A + \frac{B}{\lambda_n^2}$$
Solution by Sreeraj P, M.Sc Physics