A proton is projected at a stationary 6238 Ni aluminum target. The proton momentarily comes to a
halt at a distance from the center of an aluminum nucleus, equal to twice the nuclear radius. Assume that the nucleus retains its spherical shape and that the nuclear force on the proton is negligible. The initial kinetic energy of the proton, in MeV, is closest to:
To solve this problem, we need to consider the forces acting on the proton when it comes to a halt at a distance equal to twice the nuclear radius from the center of the aluminum nucleus.
Given that the nuclear force on the proton is negligible, we can assume that the only force acting on the proton is the electrostatic force due to the repulsion between the proton and the positively charged nucleus of the aluminum atom.
At the point where the proton momentarily comes to a halt, the electrostatic force must equal the initial kinetic energy of the proton.
The electrostatic force between two charged particles can be calculated using Coulomb's law:
F = k * (q1 * q2) / r^2
Where F is the force, k is the electrostatic constant (8.99 x 10^9 N*m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.
In this case, q1 is the charge of the proton (e = 1.6 x 10^-19 C), q2 is the charge of the aluminum nucleus (Z * e), and r is twice the nuclear radius (2 * R).
Since the aluminum nucleus has Z protons, its charge is Z * e.
Setting the electrostatic force equal to the initial kinetic energy of the proton:
k * (q1 * q2) / r^2 = (1/2) * mv^2
Where m is the mass of the proton and v is its initial velocity.
Plugging in the values:
(8.99 x 10^9 N*m^2/C^2) * (1.6 x 10^-19 C) * (Z * e) / (2 * R)^2 = (1/2) * m * v^2
Simplifying:
(Z * e)^2 / (4 * R^2) = (1/2) * m * v^2
To find the initial kinetic energy of the proton, we need to calculate the velocity of the proton using the given information.
The distance traveled by the proton is twice the nuclear radius (2 * R), so we can equate it to the displacement formula:
2 * R = (1/2) * a * t^2
Since the proton comes to a halt, its final velocity is 0, and we can use the following equation to find the time:
v = u + at
Plugging in the values:
0 = v0 + a * t
Since the proton starts from rest, v0 is 0:
0 = 0 + a * t
Simplifying:
a * t = 0
Since the proton momentarily comes to a halt, the acceleration is 0.
Therefore, the displacement formula becomes:
2 * R = 0
This indicates that the proton does not come to a halt twice the nuclear radius away from the center of the nucleus. It will actually continue to move beyond that point until it comes to a halt at a greater distance from the nucleus.
Therefore, the answer to the question as stated is that the proton does not momentarily come to a halt at a distance from the center of the aluminum nucleus equal to twice the nuclear radius.
To solve this problem, we need to consider the balance of forces acting on the proton when it comes to a halt at a distance equal to twice the nuclear radius. We can then use the equations of motion to determine its initial kinetic energy.
Let's start by considering the forces acting on the proton. At this distance, the only significant force acting on the proton is the electrostatic force between the proton and the aluminum nucleus. We can calculate this force using Coulomb's law:
F = (k * q1 * q2) / r^2,
where F is the electrostatic force, k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges of the proton and the nucleus (which are equal in magnitude but opposite in sign), and r is the distance between them.
Since the proton comes to a halt, the electrostatic force must be equal in magnitude and opposite in direction to the initial kinetic energy of the proton. Therefore, we can equate the electrostatic force to the initial kinetic energy of the proton:
(k * q1 * q2) / r^2 = KE,
where KE is the initial kinetic energy of the proton, and the right-hand side is just a constant value.
Now, let's substitute the appropriate values into the equation. The charge of the proton q1 is +e, where e is the elementary charge (1.602 x 10^-19 C). The charge of the nucleus q2 is -e. The distance r is given as twice the nuclear radius, so r = 2 * Rn, where Rn is the radius of the nucleus.
Substituting these values, we have:
(k * e * e) / (2 * Rn)^2 = KE.
Now, let's solve the equation for KE:
KE = (k * e^2) / (4 * Rn^2).
We can plug in the values for k (8.99 x 10^9 N m^2/C^2), e (1.602 x 10^-19 C), and Rn (nuclear radius) to get the final answer.
Unfortunately, the value of the nuclear radius of aluminum (6238 Ni) is not provided, so we cannot calculate the exact value of the initial kinetic energy of the proton. To get the closest estimate, you would need to find the nuclear radius of aluminum and substitute it into the equation above.