Obtain an expression for the frequency of radiations emitted when a hydrogen atom de-excites from level n to level (n – 1). For large n, show that the frequency equals the classical frequency of revolution of the electron in the orbit.

Solution

The frequency v of the emitted radiation when a hydrogen atom de-excites from level n to level (n — 1) is given by,

E = hv = E_{2} - E_{1} v = \frac{1}{2} \frac{{mc}^{2} α^{2}}{h} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] where α = \frac{2 {πKe}^{2}}{ch} = fine structure constant v = \frac{1}{2} \frac{{mc}^{2} α^{2}}{h} [\frac{1}{(n - 1)^{2}} - \frac{1}{n^{2}}] = \frac{{mc}^{2} α^{2}}{2 h} [\frac{n^{2} - (n - 1)^{2}}{n^{2} (n - 1)^{2}}] = \frac{{mc}^{2} α^{2} [(n + n - 1) (n - n + 1)]}{2 {hn}^{2} (n - 1)^{2}} v = \frac{{mc}^{2} α^{2} (2 n - 1)}{2 h n^{2} (n - 1)^{2}} .

For large n, (2n - 1)

\approx

2n, and

(n - 1)

\approx

n

Therefore,

v = \frac{{mc}^{2} α^{2} . 2 n}{2 h n^{2} . n^{2}} = \frac{{mc}^{2} α^{2}}{{hn}^{3}}

On putting,

α = \frac{2 π {Ke}^{2}}{ch}, we get v = \frac{{mc}^{2}}{{hn}^{3}} . \frac{4 π^{2} K^{2} e^{4}}{c^{2} h^{2}}

i.e.,

v = \frac{4 π^{2} {mK}^{2} e^{4}}{n^{3} h^{3}}

... (1)

In Bohr's atomic model, velocity of electron in nth orbit is,

v = \frac{nh}{2 π mr}

; and

Radius of nth orbit is,

r = \frac{n^{2} h^{2}}{4 π^{2} {mKe}^{2}} (∵ Z = 1)

∴

Frequency of revolution of electron is,

v = \frac{v}{2 πr} = \frac{nh}{2 π mr} (\frac{4 π^{2} {mKe}^{2}}{2 π . n^{2} h^{2}})

v = \frac{{Ke}^{2}}{nh . r} = \frac{{Ke}^{2}}{nh} (\frac{4 π^{2} {mKe}^{2}}{n^{2} h^{2}}) v = \frac{4 π^{2} {mK}^{2} e^{4}}{n^{3} h^{3}}

which is same as (1).

Hence proved that, for large values of n, classical frequency of revolution of electron in nth orbit is the same as the frequency of radiation emitted when hydrogen atom de-excites from level (n) to level (n – 1).

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Obtain an expression for the frequency of radiations emitted when a hydrogen atom de-excites from level n to level (n – 1). For large n, show that the frequency equals the classical frequency of revolution of the electron in the orbit.

Some More Questions From Wave Optics Chapter

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