Question

A-Calculate the speed of a proton after it accelerates from rest through a potential difference of 220V .

B-Calculate the speed of an electron after it accelerates from rest through a potential difference of 220V .

Answer 1

Potential difference = work done per unit charge
Therefore work done = potential difference * charge
Charge on a proton = 1.6 * 10^-19 C
Potential difference = 220 V
Therefore
work W = 220 * 1.6 * 10^-19 J = 351.6 * 10^-19 J = 3.516 * 10^-17 J
Let speed of proton = v
Initial kinetic energy = 0 because the proton is initially at rest.
Final kinetic energy = 1/2 mv^2
where m = mass of proton
Change in K.E. = 1/2 mv^2 - 0 = 1/2 mv^2
By work energy theorem
change in K.E. = work done
Or 1/2 mv^2 = W
Or v = sqrt(2W/m)------------------------------...
Or v = sqrt{2 * 3.516 * 10^-17/(1.67 * 10^-27)}
= sqrt{(2 * 3.516/1.67) * 10^(-17 + 27)}
= sqrt(4.21 * 10^10)
= 2.05 * 10^5 m/s

Work done = potential difference * charge
P.E. is the same i.e. 220 eV. Both proton and electron have the same magnitude of charge (only sign is different). Therefore, magnitude of the work done is also the same.
Therefore W = 3.516 * 10^-17 J as calculated earlier
In equation (1), put the mass of the electron in place of m.
m = 9.1 * 10^-31 kg
v = sqrt(2W/m)
Or v = sqrt{2 * 3.516 * 10^-17/(9.1 * 10^-31)}
= sqrt{(2 * 3.516/9.1) * 10^(-17 + 31)}
= sqrt(0.7727 * 10^14)
= sqrt(77.27 * 10^12)
= 8.79 * 10^6 m/s