Here's the question I'm currently working on...
What initial kinetic energy must an α particle have if it is to approach a stationary lead nucleus to within a distance of 35.9 fm?
So I looked up a couple of things I thought we need for this problem. I have the atomic radius of lead, and I know that the charge of the particle is 2*e, where e is the elementary charge. I was thinking of approaching this using Coulomb's law, but couldn't figure out how to get from Force to Kinetic Energy! HELP!
Thank you SO MUCH!
As the positive particle approaches the positive nucleus it experiences a retarding force. The work done against this force is equal to the change in kinetic energy of the particle. In this case, the particle stops at the given distance from the nucleus, so all the Ke is used up and the particle velocity is zero at that distance. It gained potential energy of course and that will now push the particle away from the nucleus.
I suspect you have the equation for this gain in potential energy in your book but it is easy to derive.
F = k Q q/r^2 (Coulomb)
integral Work = F dr from oo to R (which is 35.9 fm)
---> k Q q (1/R) = Ke
so, q is the elementary charge times 2, and Q is the charge of lead? what is that?
and thanks, the concepts are much clearer.
The charge on a lead nucleus is the number of protons (the atomic number) times the absolute value of the charge on an electron.
To solve this problem, you're on the right track by considering Coulomb's law. Coulomb's law describes the force between two charged particles. It states that the force between two charged particles is proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
F = k * (q1 * q2) / r^2
Where F is the force between the particles, k is the electrostatic constant (approximately 9 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.
In this case, you have an α particle (helium nucleus) with a charge of 2*e (where e is the elementary charge) approaching a stationary lead nucleus. The force between them will cause the α particle to decelerate and come to rest at a distance of 35.9 fm from the lead nucleus.
To find the initial kinetic energy of the α particle, you can start by equating the work done by the electrostatic force to its initial kinetic energy:
Work = Kinetic Energy
The work done by a force is defined as the force multiplied by the distance over which it acts. In this case, the force between the α particle and lead nucleus will oppose the motion of the α particle, so the work done will be negative.
Since the α particle comes to rest at a distance of 35.9 fm from the lead nucleus, the work done by the electrostatic force can be expressed as:
Work = - (F * d)
Where d is the distance traveled by the α particle before coming to rest.
Now, using Coulomb's law for the electrostatic force, you can replace F in the equation above:
Work = - (k * (q1 * q2) / r^2) * d
Substituting the given values, the charge of the α particle is 2*e, the charge of the lead nucleus is 82*e (as lead has an atomic number of 82), and the distance traveled is 35.9 fm. Also, remember to convert fm to meters by using the conversion factor: 1 fm = 1 x 10^-15 m.
Now you have the necessary information to calculate the initial kinetic energy of the α particle using the equation:
Work = - (k * (q1 * q2) / r^2) * d = Kinetic Energy
By substituting the values, you can solve for the initial kinetic energy.