# A proton with an initial speed of 970,000 m/s is brought to rest by an electric field.?

**1** answer

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

b)

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

c)

1/2 mv^2 = V*e

Hence

its initial k.e was 4904 [eV]

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