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.

KE = 160.41 keV

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 =   ......................1

put here value

1 × =

solve we get

v = 91428571 m/s

and with radii 2.40 from equation 1

R =

2.4 × =

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×× (91428571² + 219428570² )

KE = 2.57 × J

KE = 2.57 × J × 6.242 ×

KE = 160.41 keV

kepler used the data from tycho blahe (his mentor) to construct the laws of planetary motion.