In a hydrogen atom, the electron makes the transition from (n+1)^{th}level to the n^{th} level. If n >>>1 the frequency of radiation emitted is proportional to
Option: 1 \frac{1}{n}
Option: 2 \frac{1}{n^4}
Option: 3 \frac{1}{n^3}
Option: 4 \frac{1}{n^2}

Answers (1)

An energy gap, \Delta E= hv

Here, h is Planck's constant
therefore,
Frequency= \nu

\nu=\frac{\Delta E}{h}=k\left[\frac{1}{(n)^{2}}-\frac{1}{(n+1)^{2}}\right]\Rightarrow \nu=\frac{k (2n+1)}{n^{2}(n+1)^{2}}\\ \ since \ \ n >>> 1 \\So \ (n+1)^2\approx n^2\\ \\\Rightarrow \nu \propto \frac{1}{n^{3}}

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