Calculate the de-Broglie wavelength of an ele | vedclass

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(B) According to the de-Broglie hypothesis for a stationary orbit,the circumference of the orbit is an integral multiple of the wavelength: $n \lambda

Vedclass · Accepted Answer

$2.116 \pi \ \mathring{A}$

Vedclass · Answer

(B) According to the de-Broglie hypothesis for a stationary orbit,the circumference of the orbit is an integral multiple of the wavelength: $n \lambda = 2 \pi r$ $\lambda = \frac{2 \pi r}{n}$ For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r = a_0 	imes n^2$. Substituting this into the wavelength formula: $\lambda = \frac{2 \pi (a_0 	imes n^2)}{n} = 2 \pi a_0 n$ Given $n = 2$ and $a_0 = 0.529 \ \mathring{A}$: $\lambda = 2 	imes \pi 	imes 0.529 	imes 2 \ \mathring{A}$ $\lambda = 2.116 \pi \ \mathring{A}$

Calculate the de-Broglie wavelength of an electron residing in the $2$nd Bohr orbit of a hydrogen atom. (Bohr radius,$a_0 = 0.529 \ \mathring{A}$)

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