Question

If $n \gg 1$, then the dependence of the frequency of the photon emitted as a result of the transition of an electron from ${n^{th}}$ orbit to ${\left( {n - 1} \right)^{th}}$ orbit on $n$ will be
(A) $v \propto \dfrac{1}{n}$
(B) $v \propto \dfrac{1}{{{n^2}}}$
(C) $v \propto \dfrac{1}{{{n^3}}}$
(D) $v \propto \dfrac{1}{{{n^4}}}$

Answer 1

Hint Electrons are distributed in various energy levels inside an atom. When the electron is supplied with energy from outside the electron will absorb the energy and jump into higher energy levels. But the higher energy levels are not stable. Hence the electrons cannot remain in the higher energy level for a long time. The electron will return to its original energy level emitting the excess energy as radiations. The frequency of the radiation will depend on the energy level. Here we have to find that relation.

Complete Step by step solution
Let us consider an electron making a transition from a higher level of energy ${E_n}$to a lower energy state ${E_{n - 1}}$. The frequency of the radiation emitted can be written as,
\[\nu \propto \left( {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right)\]
Where $\nu $stands for the frequency of the emitted radiation, ${n_f}$is the principal quantum number of the final state and ${n_i}$stands for the principal quantum number of the initial state.
Here the final state is the ${\left( {n - 1} \right)^{th}}$state and the initial state is the ${n^{th}}$state, applying this in the above equation we get,
$\nu \propto \left( {\dfrac{1}{{{{\left( {n - 1} \right)}^2}}} - \dfrac{1}{{{n^2}}}} \right)$
Solving the above equation we get
$v \propto \left( {\dfrac{{{n^2} - {{\left( {n - 1} \right)}^2}}}{{{n^2}{{\left( {n - 1} \right)}^2}}}} \right)$
Expanding the term ${\left( {n - 1} \right)^2}$ we get
$v \propto \left( {\dfrac{{{n^2} - \left( {{n^2} + {1^2} - 2n} \right)}}{{{{\left( {n\left( {n - 1} \right)} \right)}^2}}}} \right)$
On solving we get
$v \propto \left( {\dfrac{{2n - 1}}{{{{\left( {n\left( {n - 1} \right)} \right)}^2}}}} \right)$
In the question, it is given that, $n \gg 1$, therefore we can neglect the $1$in the above equation,
We get,
 $ v \propto \dfrac{{2n}}{{{n^4}}}$
  $ \Rightarrow v \propto \dfrac{1}{{{n^3}}}$

Therefore, The answer is: Option (C): $v \propto \dfrac{1}{{{n^3}}}$

Note
There are two types of radiation spectra that are the emission spectrum and the absorption spectrum. An emission line spectrum consists of bright lines on a dark background. The spectrum emitted by atomic hydrogen is an example of an emission spectrum. When white light is passed through a gas and its spectrum is formed some dark lines appear in the spectrum. This is called the absorption spectrum.