We have enabled User registrations.
0 votes

A proton with an initial speed of 970,000 m/s is brought to rest by an electric field. (a) Did the proton move into a region of higher potential or lower potential? higher lower (b) What was the potential difference that stopped the proton? V (c) What was the initial kinetic energy of the proton

1 Answer

0 votes
selected by
Best answer

Protons move by themselves from higher to lower potential.

Since the proton is stopped by the potential, it was moving toward a high potential.

1/2 mv^2 = V*e
m =1.67 × 10-27 kg
e =1.602 e-19 C

V = 1/2 mv^2 /e = 0.5*1.67e-27*970000*970000/1.602 e-19

V = 4904 V

1/2 mv^2 = V*e
its initial k.e was 4904 [eV]

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

Related questions