Question

A proton has an initial speed of 5.0×10⁵ m/sm/s .

a) What potential difference is required to bring the proton to rest?

b) What potential difference is required to reduce the initial speed of the proton by a factor of 2?

c) What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 2?

Answer 1

We use

(K+U) = constant

where K is kinetic energy = 1/2*m*v²

U is potential energy = V*q

where V is potential difference

So (K + V*q)i = (K+V*q)f

a) Here Kf = 0...

1/2*m*vi² = q*(Vf-Vi)...So
(Vf-Vi) = 1/2*m*v²/q = 1/2*1.67x10^-27kg*(5x10^5m/s)^2/1.60x10^-19C

∆V= 1304.7V



b) Here Kf = 1/2*m*(vi/2)^2

So. From (K+V*q)i = (K + V*q)f we have 1/2*m*vi^2 = q*(Vf-Vi) + 1/2(m*(vi/2)^2)

So (Vf-Vi) = 1/2*m*(vi^2 - vi^2/4)/q =
1/2*1.67x10^-27*((5x10⁵)²-(5x10^5)²/4)/1.60x10^-19

∆V= 978.5V



c) Here Kf = 1/2Ki
so from (K+V*q)i = (K + V*q)f we have1/2*m*vi²= q*(Vf-Vi) + 1/2(1/2*m*vi²)

Therefore (Vf-Vi) = 1/4*m*v^2/q = 1/4*1.67x10^-27*(5x10⁵)²/1.60x10^-19

∆V= 652.3V