Magnetic moment due to the motion of the elec | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnetic moment $M$ of an electron moving in a circular orbit of radius $R$ with speed $v$ is given by $M = iA$,where $i$ is the equivalent cu

Vedclass · Accepted Answer

$n$

Vedclass · Answer

(B) The magnetic moment $M$ of an electron moving in a circular orbit of radius $R$ with speed $v$ is given by $M = iA$,where $i$ is the equivalent current and $A$ is the area of the orbit.The current $i$ is defined as the charge $e$ passing per unit time $T$,so $i = \frac{e}{T} = \frac{e \omega}{2 \pi} = \frac{ev}{2 \pi R}$.The area of the orbit is $A = \pi R^2$.Thus,the magnetic moment is $M = iA = \left( \frac{ev}{2 \pi R} \right) (\pi R^2) = \frac{evR}{2}$.Multiplying and dividing by the mass of the electron $m$,we get $M = \frac{e(mvr)}{2m} = \frac{eL}{2m}$,where $L = mvr$ is the orbital angular momentum of the electron.According to Bohr's quantization postulate,the angular momentum $L$ of an electron in the $n$th orbit is $L = \frac{nh}{2 \pi}$.Substituting this into the expression for $M$,we get $M = \frac{e}{2m} \left( \frac{nh}{2 \pi} \right) = \left( \frac{eh}{4 \pi m} \right) n$.Since $e$,$h$,and $m$ are constants,it follows that $M \propto n$.Therefore,the correct option is $B$.

Magnetic moment due to the motion of the electron in the $n$th energy state of a hydrogen atom is proportional to ..........

Similar Questions

Kinetic energy of the electron in a hydrogen atom is $\frac{e^{2}}{8 \pi \varepsilon_{0} r}$. Then its potential energy is . . . . . . .

How is the linear velocity $v$ of an electron in a $Bohr$ orbit related to its principal quantum number $n$?

The maximum wavelength that a sample of hydrogen atoms can absorb is

In a Bohr atom,the total energy of an electron in the $n$-th allowed orbit is $E_n$ and its angular momentum is $J_n$. Then:

The wavelength of the energy emitted when an electron transitions from the fourth orbit to the second orbit in a hydrogen atom is $20.397 \, cm$. The wavelength of the energy for the same transition in $He^+$ is .......... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module