Question

In a hydrogen atom the electron makes a transition from $(n+1{)}^{\text{th }}$ level to the $n^{\text{th }}$ level. If $n>>1,$ the frequency of radiation emitted is proportional to :

• A. $\frac{1}{n^4}$
• B. $\frac{1}{n^3}$
• C. $\frac{1}{n^2}$
• D. $\frac{1}{n}$

Answer 1

Answer: (B)

In hydrogen atom,
$E_n=\frac{-E_0}{n^2}$
Where $E_0$ is Ionisation Energy of $H$
$\to$ For transition from $(n+1)$ to $n$, the energy of emitted radiation is equal to the difference in energies of levels.
$\Delta E=E_{n+1}-E_n$
$\Delta E=E_0\left(\frac{1}{n^2}-\frac{1}{(n+1{)}^2}\right)$
$\Delta E=hv=E_0\left(\frac{(n+1{)}^2-n^2}{n^2(n+1{)}^2}\right)$
$hv=E_0\left[\frac{2n+1}{n^4{\left(1+\frac{1}{n}\right)}^2}\right]$
$hv=E_0\left[\frac{n\left(2+\frac{1}{n}\right)}{n^4{\left(1+\frac{1}{n}\right)}^2}\right]$
Since $n>>>1$
Hence, $\frac{1}{n}\simeq 0$
$hv=E_0\left[\frac{2}{n^3}\right]$
$v\alpha \frac{1}{n^3}$