A bare helium nucleus has two positive charges and a mass of

A bare helium nucleus has two positive charges and a mass of 6.64 times 10^-27 kg. Calculate its kinetic energy in joules at 2.5 % of the speed of light. What is this in electron volts? Write an expression for the voltage needed to obtain this energy?

Solution

a)

Given

V=2.5%c

since speed of light c=3*10⁸ m/s

V=(2.5/100)*(3*10⁸) =7.5*10⁶ m/s

Kinetic energy

KE=(1/2)mV² =(1/2)*(6.64*10^-27)(7.5*10⁶)²

KE=1.8675*10^-13 Joules

b)

In electron volts

KE=(1.8675*10^-13)/(1.6*10^-19)

KE=1.167*10⁶ eV or 1.167 MeV

c)

Since

KE=qV

V=KE/q

d)

Charge Q=2e =2*(1.6*10^-19)

Q=3.2*10^-19

V=(1.8675*10^-13)/(3.2*10^-19)

V=5.836*10⁵ Volts