SATHEE: magnetism-and-matter Question 7

magnetism-and-matter Question 7

Question: Q. 1. An electron of mass $m_{e}$ revolves around a nucleus of charge $+ Z e$ . Show that it behaves like a tiny magnetic dipole. Hence, prove that the magnetic moment associated with it is expressed as $\vec{μ} = - \frac{e}{2 m_{e}} \vec{L}$ , where $\vec{L}$ is the orbital angular momentum of the electron. Give the significance of negative sign.

U] [Delhi II 2017, 2014]

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Solution:

Ans. (i) Behaviour of revolving electron as a tiny magnetic dipole.

(ii) Proof of the relation $\vec{μ} = - \frac{e}{2 m_{e}} \vec{L}$

(iii) Significance of negative sign Electron, in circular motion around the nucleus constitutes a current loop which behaves like a magnetic dipole.

Current associated with the revolving electron : $4$

Magnetic moment of the loop,

$\begin{aligned} μ & = I A = \frac{e v}{2 π r} π r^{2} & = \frac{e v r}{2} = \frac{e \cdot m_{e} v r}{2 m_{e}} \end{aligned}$

Orbital angular momentum of the electron,

-ve sign signifies that the angular momentum of the revolving electron is opposite in direction to the magnetic moment associated with it.

[CBSE Marking Scheme 2017]

Magnetism And Matter

magnetism-and-matter Question 7

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Magnetism And Matter

magnetism-and-matter Question 7

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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