The de-Broglie wavelength of the electron in | vedclass

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(D) In the ground state of a hydrogen atom,the electron revolves in the first Bohr orbit $(n=1)$.According to Bohr's quantization condition,the circum

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$3.33$

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(D) In the ground state of a hydrogen atom,the electron revolves in the first Bohr orbit $(n=1)$.According to Bohr's quantization condition,the circumference of the orbit is equal to an integral multiple of the de-Broglie wavelength: $2\pi r = n\lambda$.For the ground state,$n=1$,so the circumference is equal to the de-Broglie wavelength: $\lambda = 2\pi r$.The radius of the first Bohr orbit is $r = 0.53 \ \mathring{A}$.Substituting the values: $\lambda = 2 	imes 3.1416 	imes 0.53 \ \mathring{A} \approx 3.33 \ \mathring{A}$.

The de-Broglie wavelength of the electron in the ground state of a hydrogen atom is ........ $\mathring{A}$.

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