The ratio of angular momentum of an electron | vedclass

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Question

(A) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{	ext{th}}$ orbit is given by $L = \frac{nh}{2\pi}$.The velocity

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$n^2$

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(A) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{	ext{th}}$ orbit is given by $L = \frac{nh}{2\pi}$.The velocity $v$ of an electron in the $n^{	ext{th}}$ orbit is given by $v = \frac{v_0}{n}$,where $v_0$ is a constant.We need to find the ratio $\frac{L}{v}$.Substituting the expressions,we get $\frac{L}{v} = \frac{nh/2\pi}{v_0/n}$.Simplifying this,we get $\frac{L}{v} = \frac{h}{2\pi v_0} \cdot n^2$.Since $h$,$\pi$,and $v_0$ are constants,the ratio $\frac{L}{v}$ is proportional to $n^2$.

The ratio of angular momentum of an electron in $n^{\text{th}}$ orbit of a hydrogen atom to the velocity of the electron in the $n^{\text{th}}$ orbit is proportional to

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