Question

A proton and an alpha particle are released from rest when they are 0.225 nm apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton.

A) Find the maximum speed of proton.

B) Find the maximum speed of alpha particle.

C) Find the maximum acceleration of proton.

D) Find the maximum acceleration of alpha particle.

Answer 1

A] Maximum speed of alpha be v, maximum speed of proton will be 4v by momentum conservation,

Final KE= initial PE

K*q*2q/r^2 = 0.5*(4m)*v^2 + 0.5*m*(4v)^2

9e9*2*1.6e-19^2/0.225e-9^2 = 0.5*(4*9.1e-27)*v^2 + 0.5*9.1e-27*16*v^2

v = 11209 m/s

Maximum speed of proton = 4*v = 44836 m/s

B] maximum speed of alpha =  v = 11209 m/s

C] maximum acceleration of proton= F/m = kq1q2/(r^2*m) = 9e9*2*1.6e-19^2/(0.225e-9^2*1.63e-27)

= 5.58*10^18 m/s^2

D] maximum acceleration of alpha= F/2m = kq1q2/(r^2*m) = 9e9*2*1.6e-19^2/(2*0.225e-9^2*1.63e-27)

= 2.79*10^18 m/s^2