Question

According to Bohr’s model of hydrogen atom, relation between principal quantum number ($n$) and radius of stable orbit is

A

$r$ Is proportional to $n$

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B

$r$ is proportional to ${n}^{2}$

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C

$r$ is proportional to ${n}^{-2}$

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D

$r$ is proportional to ${n}^{-1}$

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Solution

The correct option is A $r$ Is proportional to $n$The explanation for the correct option:Bohr model of a hydrogen atom:It was proposed by Neil Bohr in 1913.Electrons orbit the nucleus in atomic shells.The atom is held together by the electrostatic force between the negatively charged electron and positively charged nucleus.It describes the structure of hydrogen in terms of energy levels.Each shell is associated with the principal quantum number $n$.Explanation:Angular momentum is a property of mass in motion around a fixed axis that is conserved in a closed domain.Electron angular momentum about the nucleus is an integer multiple of $\frac{\mathrm{h}}{2\mathrm{\pi }}$, which is called reduced Planck's constant, where $\mathrm{h}$ is Planck's constant.$\mathrm{ħ}=\frac{\mathrm{h}}{2\mathrm{\pi }}$So, angular momentum, $L=mvr=\frac{n\mathrm{h}}{2\mathrm{\pi }}$where $m$ is the mass of the electron, $v$ is its velocity, and $r$ is its distance from the axis of rotation.The reason why it is an integral multiple of reduced Planck's constant is that at the integral values of reduced Planck's constant, the de Broglie wavelength of the electron is such that when the electron makes a complete revolution around the nucleus, the displacement of the electron from its mean position is same. Thus it does not lose momentum. Hence, $r$ is proportional to $n$.Hence, option (A) is the correct option.

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