Question

A proton, a deuteron (a hydrogen nucleus containing one proton and one neutron), and an alpha particle (a helium nucleus consisting of two protons and two neutrons) initially at rest are all accelerated through the same distance in the uniform electric field created by a very large charged plate. Compare their final (a) kinetic energies, (b) momentum magnitudes, (c) speeds, and (d) time intervals needed to cover the distance. (e) How do the changes in electric potential energy compare?

Answer 1

Answer: the answers are shown in the explanations below

Explanation: A proton carries a single positive charge q (proton is positive), a deuterium carries 2q (two protons) while an alpha particle carries 2q also (two protons)

The proton will have a unit subatomic mass m, the deuterium 2m (contains one proton, one neutron), and the alpha particle 4m (has two protons, two neutron).

The electric force on a particle with charge q moving through a charged plate of charge Q is

Fe = kQq /r^2

K is a constant, r is their separating distance.

So,

For the proton Fe = kQq/r^2

For deuterium Fe = kQq/r^2

For alpha part. Fe = 2kQq/r^2

Word done in moving the particles through the electric field is the kinetic energy of the particles according to energy conservation.

Work done = Fe * r (force times distance) = 0.5mv^2 (half of the mass times Square of their velocity).

For proton kE = kQq/r

Deuterium KE = kQq/r

Alpha part. KE = 2kQq/r

This means that the kinetic energy of the proton is equal to that of the deuterium and is equal to half the kinetic energy of the alpha part.

Since the kinetic energy of the particles is equal to the word done in moving this particles through the field, KE = W

For proton kQq/r = 0.5mv^2

V = square root of 2kQq/mr

Let us put square root of kQq/mr = b

Therefore V = 2^0.5 * w = 1.4b

Using the same transposition and method, velocities of deuterium and alpha part. respectively are;

V = b.

and

V = b

This means that the velocity of the proton is 1.4 times the velocity of the deuterium and the velocity of the alpha part.

For the momentum part, recall that momentum is the product mass and velocity mv

For proton momentum = m * 1.41b

= 1.41mb

For deuterium = 2m * b

= 2mb

For alpha part. = 4m * b

= 4mb

We can see that the momentum of the alpha particle is twice the momentum of the deuterium, and this has a magnitude of 4. The momentum of the proton has a magnitude of 1.4.

From the magnitude of the velocity above we can see that equal time will be taken by the alpha part and the deuterium. The proton will take a time 1.4 times shorter that that of the deuterium and the alpha part.