Question

Review. One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.00Cm and 2.40cm . The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0440T . Determine the energy (in keV ) of the incident electron.

Answer 1

The energy of the incident electron in keV is 5.51× 10—²² keV.

Capacity or ability to do work is termed energy. Energy exists in various forms. It can be in the form of mechanical, potential, thermal, kinetic, and many other forms. Joules is the SI of the energy. The potential energy of a system is defined as the energy stored due to its position. While kinetic energy is defined as the energy

stored due to its position.

Radius1 of the trajectory = 1.00 cm

Radius2 of the trajectory = 2.4 0 cm

The magnitude of the uniform magnetic field = 0.0440 T

A Uniform magnetic field is perpendicular to the trajectories.

The velocity of the incident electron is,

$v = \frac{eBR}{m}$

The energy of the incident electron is,

$K = \frac{1}{2} mv ^{2}$

$K = \frac{1}{2} \frac{(eBR)}{m}^{2}$

The energy in keV is,

1-kilo electron volt = 1000 electron volts

1 keV = 1000 eV

$1 \: eV = \frac{1}{1000} \: keV$

$= \frac{ K}{1000}$

$= \frac{eB ^{2} R ^{2} }{2000m}$

$= 5.51 \times 10 ^{ - 22} \: keV$

Therefore, the energy of the incident electron in keV is 5.51× 10—²² keV.

To know more about electrons, refer to the below link:

brainly.com/question/1255220

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