Question

A gyromagnetic ratio of the electron revolving in a circular orbit of hydrogen atom is $8.8\times {{10}^{10}}CK{{g}^{-1}}$. What is the mass of the electron? Given charge of the electron is $1.6\times {{10}^{-19}}C$.
$\begin{aligned} & \text{A}\text{. 1}\times \text{1}{{\text{0}}^{-29}}Kg \\\ & \text{B}\text{. 0}\text{.11}\times \text{1}{{\text{0}}^{-29}}Kg \\\ & \text{C}\text{. 1}\text{.1}\times \text{1}{{\text{0}}^{-29}}Kg \\\ & \text{D}\text{. }\dfrac{1}{11}\times \text{1}{{\text{0}}^{-29}}Kg \\\ \end{aligned}$

Answer 1

For a revolving electron in circular orbit, magnetic moment and angular momentum is associated with it. These values can be found out by applying general equations of uniform circular motion and charge-current relation of a particle. For calculating Gyromagnetic ratio, the magnetic moment of a particle is divided by the angular momentum associated with its revolution.

Formula used:
Magnetic moment of electron $\mu =\dfrac{evr}{2}$
Angular moment of electron $L=mvr$
Gyromagnetic ratio of electron $\gamma =\dfrac{e}{2m}$

Complete step by step answer:
Let us consider an electron is revolving around in a circular orbit of radius $r$ with velocity $v$. The mass of an electron is $m$ and the charge on the electron is $e$, both of which are constant values.

The time period $T$ of the electron’s circular orbit is given as:

$\begin{aligned} & T=\dfrac{\text{Circumference}}{\text{velocity}} \\\ & T=\dfrac{2\pi r}{v} \\\ \end{aligned}$

The current $i$ due to the motion of the electron is the charge flowing through that time period,

$\begin{aligned} & i=\dfrac{\text{Charge}}{\text{time}} \\\ & i=\dfrac{-e}{\dfrac{2\pi r}{v}}=\dfrac{-ev}{2\pi r} \\\ \end{aligned}$

The current is generated in the opposite direction of the movement of the electron as the electron is a negatively charged particle.
The magnetic moment due to a current loop of current$i$ enclosing an area $A$ is given by:
$\mu =iA$
The magnetic moment of an electron:

$\begin{aligned} & \mu =\dfrac{\dfrac{-ev}{2\pi r}}{\pi {{r}^{2}}}=\dfrac{-ev}{2\pi r}\times \pi {{r}^{2}} \\\ & \mu =\dfrac{-evr}{2} \\\ \end{aligned}$

Let’s divide and multiply the above equation by the mass of the electron,
$\mu =\dfrac{-e\left( mvr \right)}{2m}$
We know that the angular momentum L of a particle is given by:
$L=mvr$
Or,
$L=\dfrac{\mu \times 2m}{-e}$
Or,
$\mu =\left( \dfrac{-e}{2m} \right)L$

We are given that the gyromagnetic ratio of an electron revolving in a circular orbit of hydrogen atom is $8.8\times {{10}^{10}}CK{{g}^{-1}}$ and we have to calculate the mass of revolving electron.
For a revolving electron, the magnitude of magnetic moment is given by,

$\mu =\dfrac{evr}{2}$
Where,
$e$ is the charge on an electron
$v$ is the velocity of revolving electron
$r$ is the radius of the circular orbit
Angular momentum associated with revolving electron is given by,
$L=mvr$

Where,
$m$ is the mass of electron
$v$ is the velocity of revolving electron
$r$ is the radius of circular orbit
Gyromagnetic ratio is expressed as the ratio of the magnetic moment of the particle to its angular momentum. It is symbolized by $\gamma$.
$\gamma =\dfrac{\mu }{L}$
For revolving electron,
$\gamma =\dfrac{\dfrac{evr}{2}}{mvr}=\dfrac{e}{2m}$

Given,
$\begin{aligned} & \gamma =8.8\times {{10}^{10}}CK{{g}^{-1}} \\\ & e=1.6\times {{10}^{-19}}C \\\ \end{aligned}$

Putting the above values in$\gamma =\dfrac{e}{2m}$
We get,

$\begin{aligned} & 8.8\times {{10}^{10}}=\dfrac{1.6\times {{10}^{-19}}}{2m} \\\ & m=\dfrac{1.6\times {{10}^{-19}}}{2\times 8.8\times {{10}^{10}}} \\\ & m=\dfrac{1}{11}\times {{10}^{-29}}Kg \\\ \end{aligned}$

The mass of the revolving electron is $\dfrac{1}{11}\times {{10}^{-29}}Kg$
Hence, the correct option is D.

Note:
Students should remember that the electric current has a direction opposite to the direction of the flow of electrons. While calculating the magnetic moment of an electron, the magnetic moment vector is taken in direction opposite to the direction of path of revolving electron since electron carries a negative charge. While calculating the gyromagnetic ratio, we can take the magnitude of the magnetic moment and angular momentum of the electron.