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Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). Also show that for large values of n, this frequency equals to classical frequency of revolution of an electron. |
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Answer» We have: frequency, \(\nu\) = \(\frac{me^4}{(4\pi)^3\epsilon_o^2(\frac{h}{2\pi})^3}\) \(\Big[\frac{1}{(n-1)^2}-\frac{1}{n^2}\Big]\) = \(\frac{me^4(2n-1)}{(4\pi)^3\epsilon_o^2(\frac{h}{2\pi})^3n^2(n-1)^2}\) For large n, \(\nu\) = \(\frac{me^4}{32\pi^3\epsilon_o^2(\frac{h}{2\pi})^3}\times\frac{1}{h^3}\) Orbital frequency, \(\nu_e=\frac{\nu}{2\pi r}\) In Bohr model, \(\nu\) = \(\frac{\frac{nh}{2\pi}}{mr}\) and r = \(\frac{4\pi\epsilon_o(\frac{h}{2\pi})^2}{me^2}\) \(\therefore\) \(\nu_e\) = \(\frac{\frac{nh}{2\pi}}{2\pi mr^2}\) = \(\frac{me^4}{32\pi^3\epsilon_o^2(\frac{h}{2\pi})^3}\times\frac{1}{n^3}\) which is same as for large n. Derivation of frequency of radiation emitted when a hydrogen atom de excites from level n to level (n-1). v = me4 (2n-1) / (4\(\pi\))3 (h/2\(\pi\))3 n2 (n-1)2 Comparing for large values of n, with classical freuency v = v / 2\(\pi\)r |
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