Question

A proton has an initial speed of 3.2×10^5 m/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*v^2 and U is potential energy = V*q where V is potential

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

a) Here Kf = 0... so 1/2*m*vi^2 = q*(Vf-Vi)...So

(Vf-Vi) = m*v^2/2q =1.67x10^-27kg*(3.2x10^5m/s)^2/(2*1.6x1^-19) = 534.4 V

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*((3.2x10^5)^2-(3.2x10^2/4))/(1.6x10^-19) = 400.8V

c) Here Kf = 1/2Ki

so from (K+V*q)i = (K + V*q)f we have1/2*m*vi^2 = q*(Vf-Vi) + 1/2(1/2*m*vi^2)

Therefore (Vf-Vi) = 1/4*m*v^2/q = 1/4*1.67x10^-27*(3.2x10^5)^2/1.60x10^-19 = 267.2V