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 $$n^{th}$$ state and the ground state respectively. Let $$\Lambda_n$$ be the wavelength of the emitted photon in the transition from the $$n^{th}$$ state to the ground state. For large n, (A, B are constants)

We begin with the hydrogen atom energy‐level formula

$$E_n=-\frac{13.6\ \text{eV}}{n^{2}},$$

where $$n=1,2,3,\dots$$ is the principal quantum number. For an electron in the $$n^{\text{th}}$$ orbit, the kinetic energy $$K_n$$ equals the magnitude of the total energy (virial theorem):

$$K_n=-E_n=\frac{13.6\ \text{eV}}{n^{2}}.$$

The de Broglie wavelength of the electron in this state is obtained from

$$\lambda_n=\frac{h}{p_n},\qquad p_n=\sqrt{2mK_n},$$

so we have

$$\lambda_n=\frac{h}{\sqrt{2m\left(\dfrac{13.6\ \text{eV}}{n^{2}}\right)}}.$$ Substituting and collecting all constants into a single symbol $$C$$, we can write

$$\lambda_n=C\,n,\qquad\text{where}\qquad C=\frac{h}{\sqrt{2m(13.6\ \text{eV})}}.$$

Thus the de Broglie wavelength is directly proportional to $$n$$.

Next, the energy of the photon emitted when the electron jumps from level $$n$$ to the ground state (level 1) is

$$\Delta E=E_1-E_n=-\frac{13.6\ \text{eV}}{1^{2}}-\Bigl(-\frac{13.6\ \text{eV}}{n^{2}}\Bigr) =13.6\ \text{eV}\left(1-\frac{1}{n^{2}}\right).$$

The wavelength $$\Lambda_n$$ of this photon follows from the Planck-Einstein relation $$E=h\nu=\dfrac{hc}{\Lambda}$$, giving

$$\Lambda_n=\frac{hc}{\Delta E} =\frac{hc}{13.6\ \text{eV}\left(1-\dfrac{1}{n^{2}}\right)}.$$ Define the constant

$$A=\frac{hc}{13.6\ \text{eV}},$$

so that

$$\Lambda_n=\frac{A}{1-\dfrac{1}{n^{2}}}.$$

For large $$n$$ the quantity $$x=\dfrac{1}{n^{2}}$$ is very small. We employ the binomial expansion for $$\dfrac{1}{1-x}$$, stated as

$$\frac{1}{1-x}=1+x+x^{2}+\cdots\qquad(|x|\ll1).$$

Retaining only the first two terms, we get

$$\Lambda_n\approx A\left(1+\frac{1}{n^{2}}\right)=A+\frac{A}{n^{2}}.$$

Now we rewrite $$1/n^{2}$$ in terms of $$\lambda_n$$. From $$\lambda_n=C\,n$$ we find

$$\frac{1}{n^{2}}=\frac{C^{2}}{\lambda_n^{2}}.$$

Substituting this into the previous expression:

$$\Lambda_n\approx A+\frac{A\,C^{2}}{\lambda_n^{2}}.$$

Finally, calling $$B=A\,C^{2}$$ (a constant), we arrive at

$$\Lambda_n\approx A+\frac{B}{\lambda_n^{2}}.$$

This matches Option B in the given list. Hence, the correct answer is Option B.