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Last updated date: 08th Dec 2023
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# Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atomic model and consider all the possible transitions of this hypothetical particle to the first excited level. The largest wavelength photon that will be emitted has wavelength $\lambda$ (given in the terms of the Rydberg constant $R$ for the hydrogen atom) is equal to:Option:(A) $\dfrac{9}{{5R}}$ (B) $\dfrac{{36}}{{5R}}$ (C) $\dfrac{{18}}{{5R}}$ (D) $\dfrac{4}{R}$

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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.

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.