Relation between angular momentum and total engery

[steely]

Homework Statement

a) Express the total energy of an electron in the Coulomb potential of proton through the electron's angular momentum L and the shortest distance a between the proton and the electron's orbit. Hint: The electron's velocity is perpendicular to it's position vector whenever it is distance a away from the proton.

b) For a fixed L, minimize the expression found in (a) with respect to a. Show that the minimum corresponds to the case of a circular orbit. State the minimum value of the total energy for fixed L.

Homework Equations

L=sqrt(l²+l)ћ
L_z=m_lћ
L=r x mv=Iω
E_n=-m_ee4/[2(4πε₀)²ћ²n²]
n > l > |m_l| > 0
KE=.5Iω²

The Attempt at a Solution

E=U+KE => E= .5L*v/a - e^2/[4πε₀a]
I have little confidence in that as a solution. Either I've stopped short or I'm going in the wrong direction, I'm not sure. I really only want help with part (a)... Once I get that far I should be handle (b) somewhat easily, I just wanted to provide context for the problem.

 

[Jhero]

a) To express the total energy of an electron in the Coulomb potential of a proton, we need to consider the potential energy and kinetic energy of the electron. The potential energy is given by the Coulomb potential, which is proportional to the inverse of the distance between the electron and the proton, and the kinetic energy is given by the electron's angular momentum.

Let's start by looking at the potential energy. The Coulomb potential is given by V = -kq1q2/r, where k is the Coulomb constant, q1 and q2 are the charges of the two particles, and r is the distance between them. In this case, q1 is the charge of the electron and q2 is the charge of the proton. We can rewrite this as V = -e^2/(4πε0r), where e is the charge of the electron and ε0 is the permittivity of free space.

Next, let's consider the kinetic energy. We know that the electron's velocity is perpendicular to its position vector when it is a distance a away from the proton. This means that the electron's angular momentum, L, is given by L = mvr, where m is the mass of the electron and v is its speed. We can rewrite this as L = mva, since the velocity is perpendicular to the position vector, which has a magnitude of a.

Now, we can combine the potential and kinetic energies to get the total energy, E. This gives us E = V + KE = -e^2/(4πε0r) + .5L^2/mva^2. However, we want to express this in terms of the electron's angular momentum, L, and the shortest distance, a, between the proton and the electron's orbit. We can do this by substituting in our expression for L, which gives us E = -e^2/(4πε0r) + .5(mva)^2/mva^2. Simplifying this, we get E = -e^2/(4πε0r) + .5mv^2, since the a's cancel out.

But we still have r in our expression, and we want to express everything in terms of a. To do this, we can use the fact that the angular momentum is constant for a given orbit, which means that L = mva = constant. This means that v = L/(ma), and