An alpha particle, the nucleus of a helium atom, is at rest at the origin of a Cartesian
coordinate system. A proton is moving with a velocity of v towards the alpha particle
from the positive x axis direction. If the proton is initially far enough away to have no
potential energy, how close does the proton get to the alpha particle? Your answer
should be in terms of v, mp (mass of a proton), e (charge of a proton).
I shall assume that the alpha particle, which has a charge of 2e, cannot move.
The proton will stop and turn around when all if its initial kinetic energy, (1/2) mp V^2, is converted to electrostatic potential energy. The electrostatic potential energy when they are a distance r apart is
2 k e^2/r
because the alpha particle has a charge of 2e and the proton has a charge of e. k is the Coulomb constant. You will have to look that up.
When the particle stops,
(1/2) mp * v^2 = 2 k e^2/r
v is the initial velocity, when they are very far apart and the electrostatic potential energy is zero.
Solve for r
r = (4ke^2)/(mpV^2)
it stops when kinetic energy is equal to electrostatic energy. therefor:
.5mpv^2= kqq/r
.5mpv^2=(2ke^2)/(mpv^2)
solve for r: (4ke^2)/(mpv^2)
To determine how close the proton gets to the alpha particle, we need to consider the forces acting on the proton and their potential energy.
The force between two charged particles can be calculated using Coulomb's law:
F = (k * q1 * q2) / r^2
Where:
F is the force between the two particles,
k is the electrostatic constant (approximately 9 × 10^9 N m^2/C^2),
q1 and q2 are the charges of the particles, and
r is the distance between the two particles.
In this case, the proton and the alpha particle both have positive charges, so they repel each other. The proton moves towards the alpha particle due to the attractive gravitational force, but we can ignore this force since it's much weaker than the electromagnetic force between the charges.
Since the proton is initially far enough away to have no potential energy, we can assume that its initial kinetic energy is equal to the work done by the electromagnetic force as it approaches the alpha particle.
The work done by the electromagnetic force can be calculated using the formula:
W = ∫(F * dr)
Where:
W is the work done,
F is the force, and
dr is the displacement.
Let's assume the proton gets closest to the alpha particle at a distance "d" from the origin. At this point, the electric potential energy is zero because the proton would need to do work against the electrostatic force to move away from the alpha particle.
We can set up the equation for the work done:
W = ∫(F * dr) = -PE
Where:
PE is the potential energy.
Since the electrostatic force is in the x direction opposite to the proton's motion, we have:
F = -k * (e * e) / d^2
The work done can be calculated as:
W = - ∫(k * (e * e) / d^2 * dx) [limits: x = ∞ to x = d]
Integrating this expression, we get:
W = k * (e * e) / d [limits: x = ∞ to x = d]
Since the initial kinetic energy is equal to the work done, we have:
(1/2) * m_p * v^2 = k * (e * e) / d
Rearranging the equation to solve for d, we get:
d = (k * (e * e)) / (m_p * v^2)
Now, let's substitute the values:
k ≈ 9 × 10^9 N m^2/C^2 (electrostatic constant)
e = 1.602 × 10^-19 C (charge of a proton)
m_p = 1.673 × 10^-27 kg (mass of a proton)
v (given velocity of the proton)
Using these values, the expression for d becomes:
d ≈ (9 × 10^9 N m^2/C^2 * (1.602 × 10^-19 C)^2) / (1.673 × 10^-27 kg * v^2)
Simplifying this expression will give you the value of how close the proton gets to the alpha particle in terms of v, mp (mass of a proton), and e (charge of a proton).