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

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(A) According to Bohr's quantization condition,the angular momentum $J_n$ of an electron in the $n^{th}$ orbit is given by $J_n = \frac{n h}{2 \pi}$.T

Vedclass · Accepted Answer

$J_n \propto \lambda_n$

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(A) According to Bohr's quantization condition,the angular momentum $J_n$ of an electron in the $n^{th}$ orbit is given by $J_n = \frac{n h}{2 \pi}$.This implies $J_n \propto n$.According to the de-Broglie hypothesis,the wavelength $\lambda_n$ is given by $\lambda_n = \frac{h}{p} = \frac{h}{mv_n}$.For the $n^{th}$ orbit,the circumference is $2 \pi r_n = n \lambda_n$.Since $r_n \propto n^2$ and $v_n \propto \frac{1}{n}$,the angular momentum $J_n = m v_n r_n$ is proportional to $n$.From the relation $2 \pi r_n = n \lambda_n$,we have $\lambda_n = \frac{2 \pi r_n}{n}$.Substituting $r_n \propto n^2$,we get $\lambda_n \propto \frac{n^2}{n} = n$.Since $J_n \propto n$ and $\lambda_n \propto n$,it follows that $J_n \propto \lambda_n$.

The de-Broglie wavelength of an electron in the $n^{th}$ Bohr orbit is $\lambda_n$ and the angular momentum is $J_n$,then:

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