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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 2 years ago
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