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

In hydrogen atom, <br><br>E_n = ${{ - {E_0}} \over {{n^2}}}$ <br><br>Where E₀ is Ionisation Energy of H. <br><br>For transition from (n + 1) to n, the energy of emitted radiation is equal to the difference in energies of levels. <br><br>$\Delta$E = E_n+1 - E_n <br><br>= $${E_0}\left( {{1 \over {{n^2}}} + {1 \over {{{\left( {n + 1} \right)}^2}}}} \right)$$ <br><br>= $${E_0}\left( {{{{{\left( {n + 1} \right)}^2} - {n^2}} \over {{n^2}{{\left( {n + 1} \right)}^2}}}} \right)$$ <br><br>= E₀$\left( {{{2n + 1} \over {{n^4}{{\left( {1 + {1 \over n}} \right)}^2}}}} \right)$ <br><br>= E₀$$\left( {{{n\left( {2 + {1 \over n}} \right)} \over {{n^4}{{\left( {1 + {1 \over n}} \right)}^2}}}} \right)$$ <br><br>Since n &gt;&gt;&gt; 1 <br><br>Hence, ${{1 \over n} \simeq 0}$ <br><br>= ${E_0}\left( {{2 \over {{n^3}}}} \right)$ <br><br>As $\Delta$E = h$\nu$ <br><br>$\therefore$ h$\nu$ = ${E_0}\left( {{2 \over {{n^3}}}} \right)$ <br><br>$\Rightarrow$ $\nu$ $\propto$ ${1 \over {{n^3}}}$