The wavelength of radiation emitted is ${\lambda _0}$​ when an electron jumps from the third to second orbit of a hydrogen atom. For the electron jumping from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be:
A. $\dfrac{{16}}{{25}}{\lambda _0} \\ $
B. $\dfrac{{20}}{{27}}{\lambda _0} \\ $
C. $\dfrac{{27}}{{20}}{\lambda _0} \\ $
D. $\dfrac{{25}}{{16}}{\lambda _0}$

Hint:As the transition is from ${n_2} = 3$ to ${n_1} = 2$ and from ${n_2} = 4$ to ${n_1} = 2$. So in both the cases Balmer series is considered as the final state corresponds to n=2. The energy of the radiation produced increases with the difference between the degrees of transition, resulting in a decrease in wavelength.

Formula used:
Rydberg formula is given by:
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Here, ${R_H}$ is the Rydberg constant, ${n_1}$ is the final orbit, ${n_2}$ is the beginning orbit, and $\lambda $ is the wavelength of the radiation

Complete step by step solution:
In order to know that when an electron transitions from a high to a low energy level in hydrogen gas, the emission spectrum manifests as discrete lines of various wavelengths. These lines, which include some of the Balmer series and are seen in the visible spectrum. The Rydberg formula can be used to determine the wavelength of light that is emitted whenever an electron moves from a higher orbit to a lower orbit:
$\dfrac{1}{\lambda } = {R_H}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$

The Balmer series refers to a group of visible lines in the spectrum of hydrogen atoms. In both cases, when an electron transitions from a high energy level $(n = 3,4,5....)$ to a lower energy level ${n_1} = 2$, it enters the second orbit, so consider $\lambda = {\lambda _0}$ for the first case. Substitute ${n_1} = 2$and ${n_2} = 3$ in the above formula, then we have:
$\dfrac{1}{{{\lambda _0}}} = {R_H}\left( {\dfrac{1}{{{{(2)}^2}}} - \dfrac{1}{{{{(3)}^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _0}}} = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _0}}} = {R_H}\left( {\dfrac{5}{{36}}} \right)\,\,\,\,...(i) \\$
Similarly, consider $\lambda = {\lambda _1}$for the second case,
Substitute ${n_1} = 2$and ${n_2} = 4$ in the above formula, then we have:
$\dfrac{1}{{{\lambda _1}}} = {R_H}\left( {\dfrac{1}{{{{(2)}^2}}} - \dfrac{1}{{{{(4)}^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _1}}} = {R_H}\left( {\dfrac{1}{4} - \dfrac{1}{{16}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _1}}} = {R_H}\left( {\dfrac{3}{{16}}} \right)\,\,\,\,...(ii) \\$
Now, from equation $(i)$ and $(ii)$, we obtain:
$\dfrac{{\dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{{R_H}\left( {\dfrac{5}{{36}}} \right)\,}}{{{R_H}\left( {\dfrac{3}{{16}}} \right)\,}} \\$
$\Rightarrow \dfrac{{\dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{5{R_H}}}{{36}} \times \dfrac{{16}}{{3{R_H}}} \\$
$\Rightarrow \dfrac{{\dfrac{1}{{{\lambda _0}}}}}{{\dfrac{1}{{{\lambda _1}}}}} = \dfrac{{20}}{{27}} \\$
$\therefore {\lambda _1} = \dfrac{{20}}{{27}}{\lambda _0} \\$

Thus, the correct option is B.

Note: It should be noted that the wavelength of the radiation released as an electron in a hydrogen atom moves from the stationary state to infinity. Also Rydberg acquired the improvised technique for determining the wavelength in the hydrogen emission spectrum after the Balmer series. The formulae used to determine the wavelength of the hydrogen spectral lines are directly generalised by the formula provided by Rydberg, according to the analysis.