An alpha particle approaches a uranium nucleus as 1e6 m/s what is the closest this particle will get to the nucleus
how do you set this problem up? thanks!
set its original KEnergy equal to the PE at the closest point.
1/2 massalpha*1E12=kq1*q2/r
solve for r
wait
but kqq/r2 is the equation for a force
not an energy
To solve this problem, you can use the concept of the electrostatic force between the alpha particle and the uranium nucleus. The electrostatic force can be given by Coulomb's law:
F = k * (q1 * q2) / r^2
Where:
F is the electrostatic force
k is the electrostatic constant (9.0 x 10^9 Nm^2/C^2)
q1 and q2 are the charges of the two particles
r is the distance between the two particles
In this case, the alpha particle has a charge of +2e (where e is the elementary charge, 1.6 x 10^-19 C), and the uranium nucleus has a charge of +92e.
Now, to find the closest distance the alpha particle will get to the uranium nucleus, we need to find the distance at which the electrostatic force between them equals the difference of their kinetic energies. Let's assume this distance is denoted as "d".
The initial kinetic energy of the alpha particle is given by KE_initial = (1/2) * m * v^2, where m is the mass of the alpha particle and v is its initial velocity. The final kinetic energy of the alpha particle at the closest point is zero, as it comes to rest.
Since no other forces are acting on the alpha particle (ignoring external influences like gravity), the work done by the electrostatic force is equal to the change in kinetic energy (ΔKE):
F * d = ΔKE
Substituting the electrostatic force and kinetic energy equations, we get:
k * (q1 * q2) / d^2 = (1/2) * m * v^2
Now, we can rearrange the equation to solve for "d":
d^2 = (2 * k * (q1 * q2)) / (m * v^2)
d = √[(2 * k * (q1 * q2)) / (m * v^2)]
Plugging in the values for the constants and charges, you can calculate the final result to find the closest distance the alpha particle will get to the uranium nucleus.