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  • 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.

Q 12.13 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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Using Bohr's model we have.

v_{n}=\frac{e^{2}}{2nh\varepsilon _{_{0}}}

r_{n}=\frac{n^{2}h^{2}\varepsilon _{_{0}}}{m_{e}\pi e^{2}}

\\E_{n}=\frac{1}{2}mv_{n}^{2}-\frac{e^{2}}{4\pi \epsilon _{0}r_{n}^{2}}\\ \\E_{n}=-\frac{me^{4}}{8n^{2}h^{2} \epsilon_{0}^{2} }

\\E_{n}-E_{n-1}=-\frac{me^{4}}{8n^{2}h^{2} \epsilon_{0}^{2} }-(-\frac{me^{4}}{8(n-1)^{2}h^{2} \epsilon_{0}^{2} })\\ \\E_{n}-E_{n-1}=-\frac{me^{4}}{8h^{2} \epsilon_{0}^{2} }[\frac{1}{n^{2}}-\frac{1}{(n-1)^{2}}]\\E_{n}-E_{n-1}=-\frac{me^{4}}{8h^{2} \epsilon_{0}^{2} }[\frac{-2n+1}{n^{2}(n-1)^{2}}]

Since n is very large 2n-1 can be taken as 2n and n-1 as n

\\E_{n}-E_{n-1}=-\frac{me^{4}}{8h^{2} \epsilon_{0}^{2} }[\frac{-2n}{n^{2}(n)^{2}}]\\ \\E_{n}-E_{n-1}=\frac{me^{4}}{4n^{3}h^{2} \epsilon_{0}^{2} }

The frequency of the emission caused by de-excitation from n to n-1 would be

\\\nu =\frac{E_{n}-E_{n-1}}{h}\\ \nu =\frac{me^{4}}{4n^{3}h^{3} \epsilon_{0}^{2} }

The classical frequency of revolution of the electron in the nth orbit is given by

\nu =\frac{v_{n}}{2\pi r_{n}}

\nu =\frac{e^{2}}{2nh\epsilon _{_{0}}}\times \frac{m_{e}\pi e^{2}}{2\pi n^{2}h^{2}\epsilon _{_{0}}}

\nu =\frac{me^{4}}{4n^{3}h^{3} \epsilon_{0}^{2} }

The above is the same as the frequency of the emission during de-excitation from n to n-1.

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Sayak

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