Question

In a hydrogen-like atom, an electron makes a transition from an energy level with the quantum number n to another with the quantum number (n−1). If n>>1, the frequency of radiation emitted is approximately proportional to:
 A. $\dfrac{1}{{{n^2}}}$
B. $\dfrac{1}{{{n^{3/2}}}}$
C. $\dfrac{1}{{{n^3}}}$
D. $\dfrac{1}{n}$

Answer 1

Hint: When an electron is revolving in an orbit it will be having some energy. Every orbit has different energy levels and when an electron transits from higher energy level orbit to lower energy level orbit energy difference between the two levels is liberated as photons.
Formula used:
$\eqalign{
  & \Delta E = hf \cr
  & {E_n}\alpha \dfrac{1}{{{n^2}}} \cr} $

Complete answer:
In the case of a hydrogen like atom, when an electron needs to go to a higher energy level from the lower energy level then it will absorb the energy whose value is the difference between the two energy levels and then it will go to a higher energy level.
When the electron becomes unstable at a higher energy level due to any reason and if it transits to the lower energy level then the case will be like the energy difference between the two energy levels will be emitted in form of photons.
So the energy emitted will be having some frequency where the relation between the energy emitted and frequency is given as
$\Delta E = hf$ where $\Delta E$ is the energy emitted which is the difference between the two energy states and ‘f’ is the frequency of the radiation and ‘h’ is the Planck's constant.
At nth orbit energy of the electron will be
${E_n}\alpha \dfrac{1}{{{n^2}}}$
At n-1 orbit energy of the electron will be
${E_{n - 1}}\alpha \dfrac{1}{{{{(n - 1)}^2}}}$
By removing the proportionality symbol we get Energy difference as
$\eqalign{
  & {E_n} - {E_{n - 1}} = \Delta E \cr
  & \Rightarrow \Delta E = k\left( {\dfrac{1}{{{{(n - 1)}^2}}} - \dfrac{1}{{{{(n)}^2}}}} \right) \cr
  & \Rightarrow \Delta E = k\left( {\dfrac{{2n - 1}}{{{{(n)}^2}{{(n - 1)}^2}}}} \right) \cr
  & \Rightarrow n > > 1 \cr
  & \Rightarrow \Delta E = k\left( {\dfrac{{2n}}{{{{(n)}^2}{{(n)}^2}}}} \right) \cr
  & \Rightarrow \Delta E\alpha \dfrac{1}{{{n^3}}} \cr} $
Since we got energy difference between the two levels we equate it to the energy emitted which will be $\Delta E = hf$
$\Delta E\alpha \dfrac{1}{{{n^3}}}$
$\eqalign{
  & \Rightarrow \Delta E = hf \cr
  & \Rightarrow hf\alpha \dfrac{1}{{{n^3}}} \cr
  & \Rightarrow f\alpha \dfrac{1}{{{n^3}}} \cr} $

Hence option C will be the answer.

Note:
The energy formula of the orbit which we have is only valid for hydrogen like atoms i.e for the atoms which have only one electron. All other terms in the energy equation include the atomic number and some constant which are constants hence we had removed them and introduced proportionality symbols.