In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle, very far from the gold foil, with an initial kinetic energy of 2.0 MeV heading directly for a gold atom (charge +79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy. Find the distance of closest approach between the alpha particle and the gold nucleus.
convert 2MeV to joules.
energy=2eVE6=2E6*1.602×10−13 J=2*1.6E-7 Joules
Now set that energy to the PE at the closest point
EPE=k2e*79e/d=8.98*2*(1.6^2)(79)(1E-26)/r
set it equal to the initial KE
2*1.6E-7*d=8.98*2*1.6^2 E-26*79)
d= 8.98*79*1.6 E -19 meters
check that.
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To find the distance of closest approach between the alpha particle and the gold nucleus, we can equate the initial kinetic energy of the alpha particle to the final electrical potential energy.
The electrical potential energy between two charged particles can be calculated using the formula:
PE = (k * |q1 * q2|) / r
Where:
PE = Electrical potential energy
k = Coulomb's constant (8.99 x 10^9 Nm^2/C^2)
q1, q2 = Charges of the particles (in this case, the charge of the alpha particle and the charge of the gold nucleus)
r = Distance between the particles
Since the alpha particle is moving towards the gold nucleus, we need to find the distance at which all of its kinetic energy is converted to electrical potential energy. At this point, its final kinetic energy will be zero.
So, we can start by equating the initial kinetic energy to the electrical potential energy and solve for the distance of closest approach.
Initial kinetic energy = Final electrical potential energy
(1/2)mv^2 = (k * |q1 * q2|) / r
Where:
m = Mass of the alpha particle
v = Initial velocity of the alpha particle
The mass of an alpha particle (4He nucleus) is approximately 6.64 x 10^-27 kg.
The charge of an alpha particle (q1) is +2e, where e is the elementary charge (1.6 x 10^-19 C).
The charge of a gold nucleus (q2) is +79e.
The initial velocity of the alpha particle can be calculated by rearranging the kinetic energy equation:
KE = (1/2)mv^2
v^2 = (2 * KE) / m
v = sqrt((2 * KE) / m)
Plugging in the values, we have
v = sqrt((2 * 2.0 MeV * 1.6 x 10^-13 J/MeV) / (6.64 x 10^-27 kg))
Now, we can substitute this value of v and the charges into the equation and solve for the distance of closest approach (r).
To find the distance of closest approach between the alpha particle and the gold nucleus, we can equate the initial kinetic energy of the alpha particle to the electrical potential energy at the point of closest approach.
The initial kinetic energy of the alpha particle can be calculated using the formula:
KE = (1/2)mv^2
Where KE is the kinetic energy, m is the mass of the alpha particle, and v is the initial velocity of the alpha particle.
Given that the initial kinetic energy is 2.0 MeV, we need to convert it to joules. Using the conversion factor 1 eV = 1.6 x 10^-19 J, we can calculate the initial kinetic energy in joules:
KE = 2.0 MeV * (1.6 x 10^-19 J/eV) = 3.2 x 10^-19 J
Now, let's consider the electrical potential energy between the alpha particle and the gold nucleus. The electrical potential energy can be calculated using the formula:
PE = k * (|q1 * q2|/r)
Where PE is the potential energy, k is the Coulomb's constant (k = 8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges of the alpha particle and the gold nucleus respectively, and r is the distance between them.
Since both the alpha particle and the gold nucleus have positive charges, their charges can be represented as +2e and +79e respectively, where e is the elementary charge (e = 1.6 x 10^-19 C). Therefore, the charges can be written as:
q1 = +2e = +2 * (1.6 x 10^-19 C) = 3.2 x 10^-19 C
q2 = +79e = +79 * (1.6 x 10^-19 C) = 1.26 x 10^-17 C
We can now set the initial kinetic energy equal to the electrical potential energy and solve for the distance of closest approach (r):
3.2 x 10^-19 J = (8.99 x 10^9 N m^2/C^2) * ((3.2 x 10^-19 C) * (1.26 x 10^-17 C))/r
Solving for r gives us:
r = [(8.99 x 10^9 N m^2/C^2) * ((3.2 x 10^-19 C) * (1.26 x 10^-17 C))] / (3.2 x 10^-19 J)
Calculating the right-hand side of the equation gives us:
r ≈ 2.83 x 10^-15 m
Therefore, the distance of closest approach between the alpha particle and the gold nucleus is approximately 2.83 x 10^-15 meters.