The figure shows standing de-Broglie waves du | vedclass

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(D) According to the de-Broglie hypothesis,the circumference of a stationary orbit must be an integral number of wavelengths:$n \lambda = 2 \pi r_n$Al

Vedclass · Accepted Answer

$\frac{36 h^{2} \varepsilon_{0}}{\pi m e^{2}}$

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(D) According to the de-Broglie hypothesis,the circumference of a stationary orbit must be an integral number of wavelengths:$n \lambda = 2 \pi r_n$Also,the angular momentum of the electron is given by:$m v_n r_n = \frac{n h}{2 \pi}$From the Bohr model,the velocity of an electron in the $n^{th}$ orbit is $v_n = \frac{e^2}{2 n h \varepsilon_0}$.Substituting this into the angular momentum equation:$m \left( \frac{e^2}{2 n h \varepsilon_0} \right) r_n = \frac{n h}{2 \pi}$$r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}$In the given figure,counting the number of standing wave loops (or wavelengths),we find $n = 6$.Therefore,the radius is:$r_n = \frac{6^2 h^2 \varepsilon_0}{\pi m e^2} = \frac{36 h^2 \varepsilon_0}{\pi m e^2}$

The figure shows standing de-Broglie waves due to the revolution of an electron in a certain orbit of a hydrogen atom. Then,the expression for the orbit radius is (All notations have their usual meanings).

Similar Questions

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