Question

A proton has an initial speed of 4.4×10^5 m/s .

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

Express your answer using two significant figures.

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

Express your answer using two significant figures.

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

Express your answer using two significant figures.

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) = 1/2*m*v^2/q

= 1/2*1.67x10^-27kg*(4.4x10^5m/s)^2/1.60×10^-19

= 1010.3 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*((4.4x10^5)^2-(4.4x10^5)^2/4)/(1.60×10^-19) = 758 V



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*(4.4x10^5)^2/(1.60x10^-19) = 505.2 V