A proton and an alpha particle are released from rest when they are 0.240 nm apa

Question

A proton and an alpha particle are released from rest when they are 0.240 nm apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton.

Part A

Find the maximum speed of proton.

Part B

Find the maximum speed of alpha particle.

Part C

Find the maximum acceleration of proton.

Part D

Find the maximum acceleration of alpha particle.

Part E

When do maximum acceleration occur, just following the release of the particles or after a very long time?

Part F

When do maximum speed occur, just following the release of the particles or after a very long time?

Explanation / Answer

A)
let

m1 is the mass of proton

m1 = 1.67*10^-27 kg

mass of alfa particle, m2 = 4*m1

v1 is the speed of proton.

v2 is the speed of alfa particle.

Apply conservation of momentum

momentum gained by proton = momentum gained by alfa particle in the opposite direction.

m1*v1 = m2*v2

m1*v1 = 4*m1*v2

==> v2 = v1/4

now apply conservation of energy

final kinetic energy = initial electric potential energy

(1/2)*m1*v1^2 + (1/2)*m2*v2^2 = k*q1*q2/d

(1/2)*m1*v1^2 + (1/2)*4*m1*(v1/4)^2 = k*q1*q2/d

(1/2)*m1*v1^2 + (1/8)*m1*v1^2 = k*q1*q2/d

0.625*m1*v1^2 = k*q1*q2/d

v1 = sqrt(k*q1*q2/(0.625*m1*d))

= sqrt(9*10^9*1.6*10^-19*2*1.6*10^-19/(0.625*1.67*10^-27*0.24))

= 1.36 m/s <<<<<----------------Answer

B) v2 = v1/4

= 1.36/4

= 0.34 m/s <<<<<----------------Answer

C) F = k*q1*q2/d^2

= 9*10^9*1.6*10^-19*2*1.6*10^-19/0.24^2

= 8*10^-27 N

a1 = F/m1

= 8*10^-27/(1.67*10^-27)

= 4.79 m/s^2 <<<<<----------------Answer

D) a2 = F/m2

= 8*10^-27/(4*1.67*10^-27)

= 1.20 m/s^2 <<<<<----------------Answer

E) just following the release of the particles

F) after a very long time

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