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Chapter 36: Q. 58 (page 1022)

What is the speed, in m/s, of a proton after being accelerated from rest through a 50×106Vpotential difference?

Short Answer

Expert verified

v=9.41×107m/s

Step by step solution

01

Given Information

We have to find the speed, in m/s, of a proton after being accelerated from rest through a 50×106Vpotential difference.

02

Simplify

Energy of the proton in the potential difference is

Ek=Vq=50×106×1.6×10-19=8.0×10-12J

The mass of the proton is m=1.67×10-27. So if the velocity of the proton is vthen we have

Ek=mc211-(v/c)2-1⇒1.67×10-27×(3×108)211-(v/3×108)2-1 ⇒v=9.41×107

So the velocity of the proton will be 9.41×107m/s.

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