Question

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 1.00 cm and 2.40 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0540 T. Determine the energy (in keV) of the incident electron.

Answer 1

KE = 160.41 keV

Step-by-step explanation:

given data

radii = 1.00 cm

radii = 2.40 cm

uniform magnetic field of magnitude = 0.0540 T

to find out

energy (in keV) of the incident electron

solution

first we find the velocity of charge with radii 1 cm then 2.40 cm by given formula

R =
`(mv)/(qB)` ......................1

put here value

1 ×
`10^(-2)` =
`(9.1*10^(-31)*v)/(1.6*10^(-19)*0.052)`

solve we get

v = 91428571 m/s

and with radii 2.40 from equation 1

R =
`(mv)/(qB)`

2.4 ×
`10^(-2)` =
`(9.1*10^(-31)*v1)/(1.6*10^(-19)*0.052)`

solve we get

v1 = 219428570 m/s

so

now we calculate energy of incident electron

KE = 0.5 m (v² + v1² )

KE = 0.5 ×9.1×
`10^(-31)`× (91428571² + 219428570² )

KE = 2.57 ×
`10^(-14)` J

KE = 2.57 ×
`10^(-14)` J × 6.242 ×
`10^(15)`

KE = 160.41 keV