What is the de-Broglie wavelength of an elect | vedclass

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Question

(A) For a given orbital,the magnetic quantum number $m$ ranges from $-l$ to $+l$. Given $m = +2$,the azimuthal quantum number $l$ must be at least $2$

Vedclass · Accepted Answer

$9.97$

Vedclass · Answer

(A) For a given orbital,the magnetic quantum number $m$ ranges from $-l$ to $+l$. Given $m = +2$,the azimuthal quantum number $l$ must be at least $2$.Since $l = n - 1$ for the outermost orbit of a shell,the minimum principal quantum number $n$ is $l + 1 = 2 + 1 = 3$.According to the Bohr-de Broglie relation,the circumference of the orbit is an integral multiple of the de-Broglie wavelength: $2\pi r = n\lambda$.The radius of the $n^{th}$ orbit is given by $r = 0.529 	imes n^2 \,\mathring{A}$.Substituting $r$ into the relation: $\lambda = \frac{2\pi r}{n} = \frac{2\pi 	imes 0.529 	imes n^2}{n} = 2\pi 	imes 0.529 	imes n$.For $n = 3$,$\lambda = 2 	imes 3.1416 	imes 0.529 	imes 3 \approx 9.97 \,\mathring{A}$.

What is the de-Broglie wavelength of an electron,in a hydrogen atom,moving in an orbit having a maximum magnetic quantum number $m = +2$ in units of $\mathring{A}$?

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