The de-Broglie wavelength of the electron in | vedclass

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

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$3.33 \mathring A$

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(D) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.For the ground state of a hydrogen atom,$n = 1$.Thus,$mvr = \frac{h}{2\pi}$,which implies $2\pi r = \frac{h}{mv}$.Since the de-Broglie wavelength is defined as $\lambda = \frac{h}{mv}$,we have $\lambda = 2\pi r$.For the ground state,the radius of the first orbit is $r = 0.53 \mathring A$.Substituting the values,we get $\lambda = 2 	imes 3.14 	imes 0.53 \mathring A = 3.33 \mathring A$.

The de-Broglie wavelength of the electron in the ground state of a hydrogen atom is

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