Question

The frequency of light emitted for the transition $n = 4$ to $n = 2$ of the $H{e^ + }$ is equal to the transition in, H atom corresponding to which of the following?
A. $n = 2$ to $n = 1$
B. $n = 3$ to $n = 2$
C. $n = 4$ to $n = 3$
D. $n = 3$ to $n = 1$

Answer 1

Bohr was able to derive an equation that matched the relationship obtained from the analysis of the spectrum of the hydrogen atom. According to his model, the frequency of the light emitted by a hydrogen atom when the electron falls from a high energy $\left( {n\; = 4} \right)$ orbit into a lower energy $\left( {n\; = 2} \right)$ orbit. The same is applicable for hydrogen-like species $H{e^ + }$ .
Formula used:
$\Rightarrow$ $\nu = (cons\tan t){Z^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$

Complete step by step answer:
Planck's equation states that the energy of a photon is proportional to its frequency i.e. $\Delta E = h\nu$ . From Bohr’s model, substituting the relationship between the frequency, wavelength and the speed of light into this equation tells us that the energy of a photon is inversely proportional to its wavelength and the inverse of the wavelength of electromagnetic radiation is directly proportional to the energy of this radiation. So, we can say that frequency is directly proportional to the energy of this radiation.
We know the formula of change in energy and this can be used to determine frequency and wavelength of the light emitted by a hydrogen atom when the electron falls from a high energy orbit into a lower energy orbit as-
$\Delta E = h\nu = \dfrac{{2{\pi ^2}m{Z^2}{e^4}{k^2}}}{{{h^2}}}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$ if we see that electron falls from ${n_2}$ level to ${n_1}$ level.
Therefore, we can calculate frequency for $H{e^ + }$ when the electron falls from $n = 4$ level to ${n_1} = 2$ level.
$\Rightarrow$ $\nu = (cons\tan t){Z^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$ where constant $= \dfrac{{2{\pi ^2}m{e^4}{k^2}}}{{{h^3}}}$
Since $[\because {Z_{H{e^ + }}} = 2]$ , we have
$\Rightarrow$ ${\nu _{H{e^ + }}} = cons\tan t(4)\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right]$
$\Rightarrow$ $= constant \times 4\left[ {\dfrac{3}{{16}}} \right] = \dfrac{3}{4} \times$ constant
Now, we will calculate the same for hydrogen having $[{Z_H} = 1]$
$\Rightarrow$ $\nu (H) = cons\tan t{(1)^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$
For (A) from ${n_2} = 2$ level to ${n_1} = 2$ level.
$\Rightarrow$ $\nu (H) = constant\left[ {\dfrac{1}{1} - \dfrac{1}{4}} \right]$
$\Rightarrow$ $= \dfrac{3}{4} \times$constant which is equal to ${\nu _{H{e^ + }}}$ .
For (B), from ${n_2} = 3$ level to ${n_1} = 2$ level, we have
$\Rightarrow$ $\nu (H) = cons\tan t\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right]$ = $\dfrac{5}{{36}} \times$ constant which is not equals to ${\nu _{H{e^ + }}}$ .
For (C), from ${n_2}$= 4 level to ${n_1}$= 3 level, we have
$\Rightarrow$ $\nu (H) = cons\tan t\left[ {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right]$ $= \dfrac{7}{{144}} \times$ constant which is not equals to ${\nu _{H{e^ + }}}$ .
For (D), from ${n_2}$= 3 level to ${n_1}$= 1 level, we have
$\Rightarrow$ $\nu (H) = cons\tan t\left[ {\dfrac{1}{1} - \dfrac{1}{9}} \right]$ $= \dfrac{8}{9} \times$ constant which is not equals to ${\nu _{H{e^ + }}}$ .
From the above calculations, we can conclude that the frequency of light emitted for the transition $n = 4$ to $n = 2$ of the $H{e^ + }$ is equal to the transition from ${n_2} = 2$ to ${n_1} = 1$ in Hydrogen atom.

Hence, the correct option is (A).

Note:
The Rydberg Constant equation was shown by Neils Bohr. In the formulae ${R_H} = \dfrac{{m{k^2}{e^4}}}{{4\pi {h^3}c}}$ for the wavenumbers of lines in atomic spectra Rydberg constant appears. It is a function of the rest mass and charge of the electron, the speed of light, and Planck’s constant. Its value is 1.09677576 $\times {10^7}$ ${m^{ - 1}}$.