The difference between the longest wavelength line of the Balmer series and shortest wavelength line of the Lyman series for a hydrogenic atom (Atomic number $Z$) equal to $\Delta \lambda $. The value of the Rydberg constant for the given atom is:
A. $\dfrac{5}{{31}} \cdot \dfrac{1}{{\Delta \lambda {Z^2}}} \\ $
B. $\dfrac{5}{{36}} \cdot \dfrac{{{z^2}}}{{\Delta \lambda }} \\ $
C. $\dfrac{{31}}{5} \cdot \dfrac{1}{{\Delta \lambda {Z^2}}} \\ $
D. None

Hint:The longest wavelength would be created by the lowest amount of energy because the relationship between wavelength and energy is inverse. All Balmer series lines have ${n_1} = 2$, hence ${n_2} = 3,4,5,...$ produces the least amount. While when an electron transitions from ${n_2} = 2,3,4.,..$ to ${n_1} = 1$, the electron's lowest energy state, the subsequent ultraviolet emission lines are known as the Lyman series, a series of transitions in the hydrogen spectrum.

Formula used:
Wavelength formula for the Balmer series and Lyman series is given by:
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Here, $\lambda $ is the wavelength, $R$ is the Rydberg constant and $Z$ is the atomic number.

Complete step by step solution:
The wave number is the total number of electromagnetic field full wave cycles that can be contained inside a linear region of one metre. It displays the wave number in reciprocal metres. For the Lyman series ${n_1} = 1$, the energy difference between the two states displaying the transition should be at its highest for the shortest wavelength in the Lyman series ${n_2} = \infty$. Substitute the values in the above formula, then we have:
$\dfrac{1}{{{\lambda _1}}} = R{Z^2}\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right) \\$
$\Rightarrow {\lambda _1} = \dfrac{1}{{R(1 - \infty ){Z^2}}}\,\,\,\,\,...(i) \\$
Similarly, we need to identify the Balmer series' longest wavelength because, as we are well aware, the relationship between wavelength and energy is inverse. As a result, in the Balmer series, the principal quantum number, which is equal to ${n_2} = 3$, produces the lowest amount of energy. The Balmer series has ${n_1} = 2$ and ${n_2} = 3$ as the lowest energy level.

Now, using the wavelength formula for a single electron species, we get:
$\dfrac{1}{{{\lambda _2}}} = R{Z^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _2}}} = R{Z^2}\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) \\$
$\Rightarrow \dfrac{1}{{{\lambda _2}}} = R{Z^2}\left( {\dfrac{5}{{36}}} \right) \\$
$\Rightarrow {\lambda _2} = \dfrac{{36}}{{5R{Z^2}}}\,\,\,\,\,\,....(ii) \\$
Subtract the equation $(ii)$ from $(i)$ to value of the Rydberg constant for the hydrogenic atom, then we obtain:
$\Delta \lambda = {\lambda _2} - {\lambda _1} \\$
$\Rightarrow \Delta \lambda = \dfrac{{36}}{{5R{Z^2}}} - \dfrac{1}{{R(1 - \infty ){Z^2}}} \\$
$\Rightarrow \Delta \lambda = \dfrac{1}{{R{Z^2}}}\left( {\dfrac{{36}}{5} - 1} \right) \\$
$\therefore R = \dfrac{1}{{\Delta \lambda {Z^2}}}\left( {\dfrac{{31}}{5}} \right) $

Thus, the correct option is C.

Note: It should be noted that the Rydberg formula has been used to determine the wavelengths of several spectral series that have been created from the separation of the atomic hydrogen emission spectrum. These spectral lines are the result of an atom's electrons switching between two different energy levels.