A proton is located at the origin, and a second proton is located on the x-axis at x = 5.58 fm (1 fm = 10-15 m).
(a) Calculate the electric potential energy associated with this configuration.
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(b) An alpha particle (charge = 2e, mass = 6.64 10-27 kg) is now placed at (x, y) = (2.79, 2.79) fm. Calculate the electric potential energy associated with this configuration.
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(c) Starting with the three particle system, find the change in electric potential energy if the alpha particle is allowed to escape to infinity while the two protons remain fixed in place. (Throughout, neglect any radiation effects.)
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(d) Use conservation of energy to calculate the speed of the alpha particle at infinity.
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(e) If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity.
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No work shown
I just need an equation to get me started, the book does not list one that is useful. The example they give uses the equation V1 = Ke(q1/r1) the problem is I don't know q.
Look up the charge on a proton!!!!
(It is the same magnitude as the charge on an electron but positive, not negative, otherwise a hydrogen atom would not be neutral!!!)
Try U = k Q1 Q2 / r
where k = 9*10^9 N m^2/C^2
By the way an alpha particle is two protons.
Move alpha from infinity to the proper distance from first proton. Then double that for total gained by moving the alpha close to the two protons. Remember potentials can be added.
To answer each of these questions, we need to use the following formula for electric potential energy:
U = k * (q1 * q2) / r
where U is the electric potential energy, k is Coulomb's constant (9 * 10^9 Nm^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between the particles.
(a) To calculate the electric potential energy associated with the configuration of two protons, we have q1 = q2 = e, where e is the elementary charge (1.6 * 10^-19 C), and r = 5.58 fm. Plugging these values into the formula:
U = (9 * 10^9 Nm^2/C^2) * [(1.6 * 10^-19 C)(1.6 * 10^-19 C) / (5.58 * 10^-15 m)]
U = 2.31 * 10^-18 J
(b) Now we need to calculate the electric potential energy for the configuration with the alpha particle at (2.79, 2.79) fm. Since the alpha particle has charge 2e, we have q1 = 2e and q2 = e. The distance between them can be calculated using the Pythagorean theorem:
r = sqrt((2.79 fm)^2 + (2.79 fm)^2)
r = 3.95 fm
Plugging these values into the formula:
U = (9 * 10^9 Nm^2/C^2) * [(2e)(e) / (3.95 * 10^-15 m)]
U = 4.89 * 10^-18 J
(c) To find the change in electric potential energy if the alpha particle escapes to infinity, we compare the initial configuration (U2) to the final configuration (U1) where the alpha particle is infinitely far away. Since the electric potential energy at infinity is zero, we have:
Change in U = U2 - U1
Change in U = 4.89 * 10^-18 J - 0
Change in U = 4.89 * 10^-18 J
(d) Using conservation of energy, the change in electric potential energy (ΔU) is equal to the change in kinetic energy (ΔK) for an isolated system. Since the alpha particle escapes to infinity, it has zero kinetic energy at infinity. Therefore:
ΔU = ΔK = 1/2 * m * v^2
Setting the change in potential energy from part (c) equal to the change in kinetic energy:
4.89 * 10^-18 J = 1/2 * (6.64 * 10^-27 kg) * v^2
Solving for v:
v^2 = (2 * 4.89 * 10^-18 J) / (6.64 * 10^-27 kg)
v = sqrt[(2 * 4.89 * 10^-18 J) / (6.64 * 10^-27 kg)]
(e) Since the alpha particle remains fixed, the change in potential energy will be equal to the change in kinetic energy of the protons at infinity. Let's denote their speed as v. The initial electric potential energy is zero since the protons are at rest. Therefore, we have:
Change in U = ΔK = 1/2 * (2m) * v^2
Setting the change in potential energy from part (c) equal to the change in kinetic energy:
4.89 * 10^-18 J = 1/2 * (2 * 6.64 * 10^-27 kg) * v^2
Solving for v:
v^2 = (2 * 4.89 * 10^-18 J) / (2 * 6.64 * 10^-27 kg)
v = sqrt[(2 * 4.89 * 10^-18 J) / (2 * 6.64 * 10^-27 kg)]
I hope this explanation helps you understand how to approach each of the questions.