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

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(D) According to the Bohr quantization condition,the circumference of the $n^{th}$ orbit is an integer multiple of the de-Broglie wavelength: $2 \pi r

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) According to the Bohr quantization condition,the circumference of the $n^{th}$ orbit is an integer multiple of the de-Broglie wavelength: $2 \pi r_n = n \lambda$.Therefore,the wavelength is given by $\lambda = \frac{2 \pi r_n}{n}$.The radius of the $n^{th}$ orbit for a hydrogen-like ion is $r_n = a_0 \frac{n^2}{Z}$,where $a_0 \approx 0.529 \ \mathring{A}$.For $He^{+1}$ ion,the atomic number $Z = 2$. For the $3^{rd}$ orbit,$n = 3$.Substituting these values into the radius formula: $r_3 = 0.529 	imes \frac{3^2}{2} = 0.529 	imes 4.5 = 2.3805 \ \mathring{A}$.Now,substitute $r_3$ and $n$ into the wavelength formula: $\lambda = \frac{2 \pi (2.3805)}{3}$.$\lambda = \frac{2 	imes 3.14159 	imes 2.3805}{3} \approx \frac{14.958}{3} \approx 4.986 \ \mathring{A}$.Rounding to the nearest integer,we get $\lambda \approx 5 \ \mathring{A}$.

The de-Broglie wavelength of an electron in the $3^{rd}$ orbit of a $He^{+1}$ ion is approximately (in $\mathring{A}$):

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