Question

In a hydrogen atom, the electron makes the transition from an energy level with quantum number $n$ to another with quantum number $(n-1)$. If $n>>1$, the frequency of radiation emitted is proportional to

• A. $\frac{1}{n^{3/2}}$
• B. $\frac{1}{n}$
• C. $\frac{1}{n^3}$
• D. $\frac{1}{n^2}$

Answer 1

The correct option is C $\frac{1}{n^3}$

The frequency of the emitted radiation is related to energy levels as,

$\Delta E=E_n-E_{n-1}=h\nu$

Here, $h$ is Planck's constant,

According to Bohr's model, the energy of the orbits in hydrogen atom is given by,

$E_n=\frac{-13.6}{n^2}\text{eV}$

$\Rightarrow \nu =\frac{\Delta E}{h}=K[\frac{1}{(n-1{)}^2}-\frac{1}{n^2}][\text{Here,}K=\frac{13.6\text{eV}}{h}]$

$\Rightarrow \nu =\frac{K(2n-1)}{(n-1{)}^2(n{)}^2}$

As, $n>>1$

$\Rightarrow \nu =\frac{2Kn}{n^4}$

$\therefore \nu \propto \frac{1}{n^3}$

Hence, option $\left(\text{D}\right)$ is correct.

Why this Question? Key concept: A photon will be emitted with frequency $\nu$ when an electron makes a transition from higher energy state to lower energy state. $\Delta E=E_{higher}-E_{lower}=h\nu$