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