If an electron in a hydrogen atom jumps from an orbit of level $n = 3$ to an orbit of level $n = 2$, emitted radiation has a frequency ____.
(R = Rydberg’s constant, C = velocity of light).
(A) $\dfrac{{3RC}}{{27}}$
(B) $\dfrac{{RC}}{{25}}$
(C) $\dfrac{{8RC}}{9}$
(D) $\dfrac{{5RC}}{{36}}$

Hint: This question is based on “Atomic Spectra and Spectral Series”. When an electron in an atom for example hydrogen, gets excited and moves from one energy level to another energy level it forms a spectrum of specific wavelengths, this spectrum is known as the “Atomic Spectra”. In order to solve this question, we use Rydberg's Formula for the Hydrogen spectral series.

Complete step by step solution:
The Rydberg’s formula is given by:
\[\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{n_2}^2}} - \dfrac{1}{{{n_1}^2}}} \right)\]
Where,
$\lambda = $ The wavelength of the emitted photon during this electron transition
$R = $ The Rydberg constant and its value $1.0974 \times {10^7}$ per meter
$Z = $ The atomic number
${n_1} = $ The initial energy orbit level
${n_2} = $ The final energy orbit level
Given: For the hydrogen atom, the atomic number of the hydrogen $Z = 1$
the initial energy orbit level ${n_1} = 3$
and the final energy orbit level ${n_2} = 2$
Putting these values into the Rydberg’s formula, we get,
$
\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{n_2}^2}} - \dfrac{1}{{{n_1}^2}}} \right)\\
\implies \dfrac{1}{\lambda } = R \times 1\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)
$
By solving this, we get,
$
\dfrac{1}{\lambda } = R\left( {\dfrac{{9 - 4}}{{36}}} \right)\\
\implies \lambda = \dfrac{{36}}{{5R}}
$
So, the value of the wavelength of the emitted photon calculated is \[\lambda = \dfrac{{36}}{{5R}}\].
Now, to determine the frequency of the emitted photon radiation we use the relationship formula between the velocity of light $\left( C \right)$ ,the frequency $\left( \nu \right)$ and the wavelength of the emitted photon $\left( \lambda \right)$
Thus, frequency of the emitted photon radiation,
$
\nu = \dfrac{C}{\lambda }\\
 = \left( {\dfrac{C}{{{\raise0.7ex\hbox{${36}$} \!{\left/
 {\vphantom {{36} {5R}}}\right.}
\!\lower0.7ex\hbox{${5R}$}}}}} \right)\\
 = \dfrac{{5RC}}{{36}}
$

Therefore, the frequency of the emitted radiation is $\dfrac{{5RC}}{{36}}$

So, the correct answer is “Option D”.

Note:
When the electron in an atom jumps from one orbit level to another orbit level there is a transition of energy between the orbits. This energy can be calculated by using the value of frequency obtained from the Rydberg’s formula in following way-
$E = h\nu $
Where, $h = $The Planck’s constant, its value is $6.63 \times {10^{ - 34}}$Js