In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number ( $\mathrm{n}-1$ ). If $\mathrm{n} \gg 1$, the frequency of radiation emitted is proportional to

(C) $\frac{1}{\mathrm{n}^{3}}$

(A) $\frac{1}{\mathrm{n}^{2}}$

JEE Main · 2013 · 07 Apr 2013 · Offline · Modern Physics

In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number (

n - 1

). If

n ≫ 1

, the frequency of radiation emitted is proportional to

(A) $\frac{1}{n ^{2}}$
(B) $\frac{1}{n ^{3/2}}$
(C) $\frac{1}{n ^{3}}$ ✓ Correct
(D) $\frac{1}{n}$

Answer: (C)

Solution

The frequency of the emitted photon tracks the energy gap between the two levels,

ν \propto \frac{1}{( n - 1 ) ^{2}} - \frac{1}{n ^{2}} = \frac{2 n - 1}{n ^{2} ( n - 1 ) ^{2}}

. When

n

is very large the numerator

2 n - 1

behaves like

2 n

and each factor in the denominator like

n^{2}

, so

ν \propto \frac{2 n}{n ^{4}} = \frac{2}{n ^{3}}

. The emitted frequency therefore falls off as

\frac{1}{n ^{3}}

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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