1. Use the Rydberg formula for a hydrogen-like atom

For emission from level $n_2$ to $n_1$,

$\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

where $R$ is the Rydberg constant and $Z$ is the atomic number.

2. Transition from M-shell to L-shell

Here,

• M-shell $\Rightarrow n_2 = 3$
• L-shell $\Rightarrow n_1 = 2$

So,

$\frac{1}{\lambda} = RZ^2\left(\frac{1}{2^2} - \frac{1}{3^2}\right)$

$\frac{1}{\lambda} = RZ^2\left(\frac{1}{4} - \frac{1}{9}\right) = RZ^2\left(\frac{5}{36}\right)$

Thus,

$\lambda = \frac{36}{5RZ^2}$

3. Transition from N-shell to L-shell

Here,

• N-shell $\Rightarrow n_2 = 4$
• L-shell $\Rightarrow n_1 = 2$

Let the new wavelength be $\lambda'$.

$\frac{1}{\lambda'} = RZ^2\left(\frac{1}{2^2} - \frac{1}{4^2}\right)$

$\frac{1}{\lambda'} = RZ^2\left(\frac{1}{4} - \frac{1}{16}\right) = RZ^2\left(\frac{3}{16}\right)$

So,

$\lambda' = \frac{16}{3RZ^2}$

4. Find the ratio $\lambda'/\lambda$

$\frac{\lambda'}{\lambda} = \frac{\frac{16}{3RZ^2}}{\frac{36}{5RZ^2}}$

$\frac{\lambda'}{\lambda} = \frac{16}{3} \cdot \frac{5}{36} = \frac{80}{108} = \frac{20}{27}$

Hence,

$\lambda' = \frac{20}{27}\lambda$

5. Match with the options

The correct option is:

D: $\dfrac{20}{27}\lambda$