1. de Broglie wavelength in the Bohr model

For hydrogen in the $n$th orbit, $2\pi r_n = n\lambda_n$ with $r_n = n^2 a_0$ So, $\lambda_n = \frac{2\pi r_n}{n} = 2\pi a_0 n$ Hence, $\lambda_n \propto n$ Therefore, $n = \frac{\lambda_n}{2\pi a_0}$

Also for the ground state, $\lambda_g = 2\pi a_0$ so equivalently, $\lambda_n = n\lambda_g$

2. Photon emitted in transition $n \to 1$

The energy levels of hydrogen are $E_n = -\frac{13.6}{n^2}\,\text{eV}$ Thus the emitted photon energy is $h\nu = E_1 - E_n = 13.6\left(1-\frac{1}{n^2}\right)\text{eV}$ Using $\frac{hc}{\Lambda_n} = 13.6\left(1-\frac{1}{n^2}\right)$ we get $\frac{1}{\Lambda_n} = R\left(1-\frac{1}{n^2}\right)$ where $R$ is the appropriate constant.

So, $\Lambda_n = \frac{1}{R\left(1-\frac{1}{n^2}\right)}$

For large $n$, $\frac{1}{1-1/n^2} \approx 1 + \frac{1}{n^2}$ Therefore, $\Lambda_n \approx \frac{1}{R} + \frac{1}{R n^2}$ This is of the form $\Lambda_n \approx A + \frac{B}{n^2}$ for constants $A,B$.

3. Express in terms of $\lambda_n$

Since $\lambda_n \propto n$ we have $n^2 \propto \lambda_n^2$ Therefore, $\frac{1}{n^2} \propto \frac{1}{\lambda_n^2}$ Hence, $\Lambda_n \approx A + \frac{B}{\lambda_n^2}$

So the correct option is: $\boxed{\text{A}}$

4. Check other options

• B: $\Lambda_n \approx A + B\lambda_n$ would grow linearly with $n$, but $\Lambda_n$ approaches a finite limit for large $n$. So false.
• C: $\Lambda_n^2 \approx A + B\lambda_n^2$ would grow like $n^2$, false.
• D: Incomplete/incorrect relation, not consistent.

Thus only A is correct.