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

\(E _{ n }=\frac{- E _{0}}{ n ^{2}}\)

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.

\(\Delta E = E _{ n +1}- E _{ n }\)

\(\Delta E = E _{0}\left(\frac{1}{ n ^{2}}-\frac{1}{( n +1)^{2}}\right)\)

\(\Delta E = h v= E _{0}\left(\frac{( n +1)^{2}- n ^{2}}{ n ^{2}( n +1)^{2}}\right)\)

\(h v= E _{0}\left[\frac{2 n +1}{ n ^{4}\left(1+\frac{1}{ n }\right)^{2}}\right]\)

\(h v= E _{0}\left[\frac{ n \left(2+\frac{1}{ n }\right)}{ n ^{4}\left(1+\frac{1}{ n }\right)^{2}}\right]\)

since \(n >>>1\)

Hence, \(\frac{1}{ n } \simeq 0\)

\(h v= E _{0}\left[\frac{2}{ n ^{3}}\right]\)

\(v \alpha \frac{1}{ n ^{3}}\)