Question

What is the radius of the path of an electron (mass $9 \times {10^{ - 31}}kg$ and charge $1.6 \times {10^{ - 19}}C$ ) moving at a speed of $3 \times {10^{ - 7}}m/s$ in a magnetic field of $6 \times {10^{ - 4}}T$ perpendicular to it? What is its frequency? Calculate its energy in keV. ( $1eV = 1.6 \times {10^{ - 19}}J$ ).

Answer 1

We start by finding the radius of the orbit in which the electron is orbiting. Then with the help of this value, we will be able to find the value of the frequency of this motion. Then we move to finding the energy of the electron. Since we have an electron which is moving, we find the kinetic energy.
Frequency can be found out using the formula, $frequency = \dfrac{\nu }{{2\pi }} = \dfrac{1}{{2\pi }}\left( {\dfrac{v}{r}} \right)$
Energy is given by, the formula, $E = \dfrac{1}{2}m{v^2}$
Charge of an electron is given by, $evB = \dfrac{{m{v^2}}}{r}$
Where $v$ is the velocity of the electron
$m$ is the mass of the electron
$r$ is the radius of the circular path of the electron.

Complete answer:
Let us start by noting down the given information.
Mass of the electron is given as $m = 9 \times {10^{ - 31}}$
Charge of an electron is given as, $q = 1.6 \times {10^{ - 19}}$
The velocity of the electron is given as, $v = 3 \times {10^{ - 7}}$
The value of one electron volt is, $1eV = 1.6 \times {10^{ - 19}}J$
The magnetic field associated with the electron is $B = 6 \times {10^{ - 4}}T$
We start by finding the radius of the path using, $evB = \dfrac{{m{v^2}}}{r}$
Taking the unknown to one side, $r = \dfrac{{m{v^2}}}{{evB}} = \dfrac{{9.1 \times {{10}^{ - 31}} \times \left( {3 \times {{10}^{ - 7}}} \right)}}{{1.6 \times {{10}^{ - 19}} \times 6 \times {{10}^{ - 4}}}} = 0.28125m$
Now that we have the value of radius of the path, we find the frequency using the formula,
$frequency = \dfrac{\nu }{{2\pi }} = \dfrac{1}{{2\pi }}\left( {\dfrac{v}{r}} \right) = \dfrac{1}{{2\pi }}\left( {\dfrac{{3 \times {{10}^{ - 7}}}}{{0.28125}}} \right) = 1.7 \times {10^7}Hz$
We move to find the value of energy. Since we have an electron moving in a path, this energy will be kinetic,
Kinetic energy is given by, $E = \dfrac{1}{2}m{v^2} = \dfrac{1}{2} \times 9.1 \times {10^{ - 31}} \times {\left( {3 \times {{10}^{ - 7}}} \right)^2} = 4.05 \times {10^{ - 6}}J$
It is specifically asked that the value should be in electron volts, which means we divide the value we got with the charge of an electron and get
$2.53 \times {10^4}eV = 25.3keV$
The energy in electron volts will be $25.3keV$ .

Note:
Frequency is defined as the rate of rotation, or the number of rotations in some unit of time. Angular frequency is the rotation rate measured in radians. We should convert the uit of the answer as required in the question.