Answer

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**Hint:**We know that wavelength emitted due to transition of electrons is predicted by using Rydberg formula. It is basically a mathematical way to represent wavelength due to the travelling of electrons between energy levels of an atom. The transition mainly occurs due to absorption of electromagnetic radiation.

**Complete answer:**

Now, let us consider Rydberg formula to solve this problem

Where the wavelength can be determined as:

$ \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_b}}} - \dfrac{1}{{{n_i}}}} \right) $ where, $ \lambda = $ wavelength, $ R = $ Rydberg constant, $ {n_b} = $ lower energy level $ {n_i} = $ higher energy level.

$ R $ is given as $ R = \dfrac{{{m_e}.{e^4}}}{{8{\varepsilon _0}c{h^3}}} $ where $ {m_e} $ is mass of electron, $ e = $ elementary charge, $ {\varepsilon _0} = $ permittivity of vacuum, $ c = $ speed of light in vacuum, $ h = $ planck's constant

It is given that electrons have the same charge so $ e $ will be constant, other elements in the formula are also constant such as $ h,{\varepsilon _0},c $ .

So we can say that Rydberg constant $ R \propto {m_e} $

It is given that $R' = 2R $

When we are talking about first excited state the electrons jump from second excited state to first and we will have to find out only one wavelength

So let us find out

For wavelength we have lower energy level two and higher energy level three which means $ {n_b} = 2,{n_i} = 3 $ and as given in question $ R' = 2R $

$

\dfrac{1}{\lambda } = 2R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) \\

= 2R\left( {\dfrac{1}{4} - \dfrac{1}{9}} \right) \\

\dfrac{1}{\lambda } = \dfrac{{5R}}{{18}} \\

\lambda = \dfrac{{18}}{{5R}} \\

$ H

We have got this value in given option

**Hence, the correct option is C.**

**Note:**

Here, $ {n_i} $ will always be greater than $ {n_b} $ . For calculating wavelengths of spectral lines in any chemical element Rydberg’s formula is used which is based upon Bohr’s atomic model which explains the atomic spectrum of hydrogen and other atoms, ions.

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