The de-Broglie wavelength of an electron in t | vedclass

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(D) According to Bohr's quantization condition,the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr_n = \frac{nh}{2\pi}$.Since t

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Equal to the circumference of the first orbit

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(D) According to Bohr's quantization condition,the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr_n = \frac{nh}{2\pi}$.Since the momentum $p = mv = \frac{h}{\lambda}$,we can substitute this into the equation:$\frac{h}{\lambda} 	imes r_n = \frac{nh}{2\pi}$.Rearranging for the wavelength $\lambda$,we get $\lambda = \frac{2\pi r_n}{n}$.For the first Bohr orbit,$n = 1$,so $\lambda = 2\pi r_1$.Since the circumference of the first orbit is $2\pi r_1$,the de-Broglie wavelength is equal to the circumference of the first orbit.

The de-Broglie wavelength of an electron in the first Bohr orbit is

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