When the electron orbiting in a hydrogen atom | vedclass

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(D) According to Bohr's postulate for the quantization of angular momentum,$mvr = \frac{nh}{2\pi}$.From the de-Broglie hypothesis,the wavelength is gi

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$\lambda \propto n$

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(D) According to Bohr's postulate for the quantization of angular momentum,$mvr = \frac{nh}{2\pi}$.From the de-Broglie hypothesis,the wavelength is given by $\lambda = \frac{h}{mv}$.Rearranging the angular momentum equation,we get $mv = \frac{nh}{2\pi r}$.Substituting this into the de-Broglie wavelength formula: $\lambda = \frac{h}{nh / (2\pi r)} = \frac{2\pi r}{n}$.For a hydrogen atom,the radius of the $n^{th}$ orbit is $r_n \propto n^2$.Substituting $r \propto n^2$ into the wavelength expression: $\lambda \propto \frac{n^2}{n} = n$.Therefore,$\lambda \propto n$.

When the electron orbiting in a hydrogen atom goes from one orbit to another orbit (principal quantum number $= n$),the de-Broglie wavelength $(\lambda)$ associated with it is related to $n$ as:

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