Question

A bare helium nucleus has two positive charges and a mass of 6.64 ✕ 10-27 kg.

(a) Calculate its kinetic energy in joules at 4.60% of the speed of light _____J

(b) What is this in electron volts? ___eV

(c) What voltage would be needed to obtain this energy? ______V

Answer 1

a)

4.6% of light speed = 4.6*3*10^8 /100=13.8*10^6

E= 0.5*m*v^2 = 0.5*(13.8*10^6)^2*6.64*10^-27 =6.322*10^-13 J

b)

1 eV = 1.6×10-19 joule

so = 6.322*10^-13 J/ 1.6×10-19= 3 951 630 ev

kinetic energy = 3 951 630 eV

c)

Voltage

since the charge is = 2 X charge of an electron
charge = 2 electrons
2 X V = 3 951 630
=> V = 3 951 630/2= 1 975 815  V