What is the de Broglie wavelength of an elect | vedclass

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

Vedclass · Accepted Answer

$3.14 	imes 10^{-8} 	imes n \ cm$

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(C) According to Bohr's postulate,the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this,we get $2\pi r = \frac{nh}{mv}$.From de Broglie's relation,the wavelength $\lambda = \frac{h}{mv}$.Substituting this into the equation,we get $2\pi r = n\lambda$,or $\lambda = \frac{2\pi r}{n}$.For a hydrogen atom,the radius of the $n^{th}$ orbit is $r_n = 0.529 	imes 10^{-8} 	imes n^2 \ cm$.Substituting $r_n$ into the expression for $\lambda$: $\lambda = \frac{2 	imes 3.1416 	imes 0.529 	imes 10^{-8} 	imes n^2}{n} = 3.32 	imes 10^{-8} 	imes n \ cm$.Since the provided options are approximations,the closest form is $3.14 	imes 10^{-8} 	imes n \ cm$.

What is the de Broglie wavelength of an electron in the $n^{th}$ Bohr orbit?

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