The Rydberg Constant R for hydrogen is:
A) $ R = - \left( {\dfrac{1}{{4\pi {\varepsilon _0}}}} \right)\dfrac{{2{\pi ^2}m{e^2}}}{{c{h^2}}} $
B) $ R = \left( {\dfrac{1}{{4\pi {\varepsilon _0}}}} \right)\dfrac{{2{\pi ^2}m{e^4}}}{{c{h^2}}} $
C) $ R = {\left( {\dfrac{1}{{4\pi {\varepsilon _0}}}} \right)^2}\dfrac{{2{\pi ^2}m{e^4}}}{{{c^2}{h^2}}} $
D) $ R = {\left( {\dfrac{1}{{4\pi {\varepsilon _0}}}} \right)^2}\dfrac{{2{\pi ^2}m{e^4}}}{{c{h^3}}} $

Answer
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Hint: The Rydberg constant is denoted by the symbol $ {R_\infty } $ for heavy metals and $ {R_H} $ for hydrogen atoms. It is named after the Swedish physicist Johannes Rydberg, and is a physical constant relating to the electromagnetic spectrum of an atom.

Complete Step By Step Answer:
The value of the Rydberg constant was first found as an empirical fitting parameter but later it was found that its value could be more accurately measured using the Bohr Model. By 2018 the value of $ {R_\infty } $ was most accurately found. This constant is used to depict the highest wavenumber of a photon that can be emitted from an atom. The hydrogen spectral series can be expressed in terms of $ {R_H} $ and the Rydberg formula.
The value of $ {R_\infty } $ can be given by the formula: $ {R_\infty } = \dfrac{{{m_e}{e^4}}}{{8\varepsilon _0^2{h^3}c}} $
Where $ {m_e} $ is the mass of the electron, e is the elementary charge, h is the planck's constant, c is the speed of light.
The value of $ {R_H} $ can be calculated by the reduced mass of the electron and can be given by the formula: $ {R_H} = {R_\infty }\dfrac{{{m_p}}}{{{m_e} + {m_p}}} \approx 1.09678 \times {10^7}{m^{ - 1}} $
Where, $ {m_e} $ is the mass of electrons and $ {m_p} $ is the mass of protons.
Therefore we know that, $ {R_\infty } = \dfrac{{{m_e}{e^4}}}{{8\varepsilon _0^2{h^3}c}} $ . On simplifying the value of $ {R_\infty } $ we get, $ R = \dfrac{{2{\pi ^2}m{e^4}{k^2}}}{{c{h^3}}} $ . Simplifying the value of k as $ 1/4\pi {\varepsilon _0} $ we get the value of R as $ R = {\left( {\dfrac{1}{{4\pi {\varepsilon _0}}}} \right)^2}\dfrac{{2{\pi ^2}m{e^4}}}{{c{h^3}}} $
The correct answer is Option (D).
Additional Information: The Rydberg formula is used to find out the wavelength emitted when an electron moves between energy levels of an atom. The Rydberg formula applicable to different elements can be given as:
 $ \mathop \nu \limits^\_ = \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) $
Where R is the Rydberg Constant.

Note:
In atomic physics, Rydberg Unit of energy, denoted by the symbol $ {R_y} $ corresponds to the energy of the photon, whose wave number is equal to the Rydberg Constant or the ionization energy of a Hydrogen atom in a simplified Bohr Model.