What is the de Broglie wavelength of an elect | vedclass

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(B) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this,we

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$\pi r$

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(B) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this,we get $2\pi r = \frac{nh}{mv}$.From de Broglie's relation,the wavelength $\lambda = \frac{h}{mv}$.Substituting this into the equation,we get $2\pi r = n\lambda$,which implies $\lambda = \frac{2\pi r}{n}$.For the $2^{nd}$ orbit,$n = 2$.Therefore,$\lambda = \frac{2\pi r}{2} = \pi r$.

What is the de Broglie wavelength of an electron in the $2^{nd}$ orbit of a hydrogen atom?

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