The de Broglie wavelength of an electron in t | vedclass

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(C) According to the Bohr's quantization condition, the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.Rear

Vedclass · Accepted Answer

$6 \pi 	imes 52.9 \ pm$

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(C) According to the Bohr's quantization condition, the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this, we get $2\pi r = \frac{nh}{mv}$.Since the de Broglie wavelength is $\lambda = \frac{h}{mv}$, we can substitute this into the equation to get $2\pi r = n\lambda$, or $\lambda = \frac{2\pi r}{n}$.For the $H$-atom, the radius of the $n^{th}$ orbit is $r_n = n^2 	imes a_0$, where $a_0 = 52.9 \ pm$.For the third orbit $(n = 3)$, $r_3 = 3^2 	imes 52.9 \ pm = 9 	imes 52.9 \ pm$.Substituting $n = 3$ and $r_3$ into the wavelength formula: $\lambda = \frac{2\pi 	imes (9 	imes 52.9 \ pm)}{3} = 6\pi 	imes 52.9 \ pm$.

The de Broglie wavelength of an electron in the third Bohr orbit of $H$-atom is

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