One electron collides elastically with a second electron initially at rest. After the collision, the radii...

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.00 cm and 2.50cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0380 T. Determine the energy (in keV) of the incident electron.

Solutions

Expert Solution

The magnetic force on electron during the trajectory is

m v ² / r = B q v

v = B q r / m

here q is charge on electron, r radius of trajectories, and m is mass of electron

The speed of each electron after collision is

and

kinetic energy of each electron is

K1 = ( 1/2) m v ₁²

K2 = ( 1/2) m v ₂²

Therefore, the total energy of the incident is,

In terms of keV,

genius_generous answered 1 year ago

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