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

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Solution

$v = \frac{m e^{4}}{(4 π)^{3} ε_{0}^{2} (h / 2 π)^{3}} [\frac{1}{(n - 1)^{2}} - \frac{1}{n^{2}}] = \frac{m e^{4} (2 n - 1)}{(4 π)^{2} ε_{0}^{2} (h / 2 π)^{3} n^{2} (n - 1)^{2}}$
For large n, $v ≅ \frac{m e^{4}}{32 π^{3} ε_{0}^{2} (h / 2 π)^{3} n^{3}}$
Orbital frequency $v_{e} = (v / 2 π r)$ .
In Bohr model, $v = \frac{n (h / 2 π)}{m r}$ , and $r = \frac{4 π ε_{0} (h / 2 π)^{2}}{m e^{2}} n^{2}$ . This gives $v_{e} = \frac{n (h / 2 π)}{2 π m r^{2}} = \frac{m e^{4}}{32 π^{3} ε_{0}^{2} (h / 2 π)^{3} n^{3}}$ which is same as for large n.

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Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

v=me4(4π)3ε20(h/2π)3[1(n−1)2−1n2]=me4(2n−1)(4π)2ε20(h/2π)3n2(n−1)2 For large n, v≅me432π3ε20(h/2π)3n3Orbital frequency ve=(v/2πr).In Bohr model, v=n(h/2π)mr, and r=4πε0(h/2π)2me2n2. This gives ve=n(h/2π)2πmr2=me432π3ε20(h/2π)3n3 which is same as for large n.

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