## Wednesday, July 14, 2010

### Solution of Quantum Mechanics Problems

Left: Solution diagram for Problem #1.
Let’s look at the solutions of previous problems in quantum mechanics:

1) An electron is in the n= 2 state of hydrogen. Find the possible values of: L, L(z) and theta.
If the angular momentum quantum number l = 2

Then: L = [l(l + 1)]^1/2 (h/2pi) = [2(2 +1)]^1/2 (h/2pi)
= [6]^1/2 (h/2pi)

and a set of allowed projections for L are obtained (see diagram ) as well as orientations. Of particular interest are the allowed values for the vertical component of the orbital angular momentum vector, which we designate: L(z). (L(z) is the projection of L along the z-axis and has discrete values, e.g. 0, h/2pi, h/pi, - h/2pi, - h/pi for the case given).[See Diagram]

L(z) = m_l(h/2pi)

Recall that m_l can range from -l to l, so for l = 2, we have:

m_l = +2, +1, 0, -1, -2

When we multiply each of these by (h/2pi) we obtain the quantizations of L(z) as depicted in the diagram.

The angle (theta) can indeed be found using the cosine relation (adjacent over the hypoteneuse) which yields (using 't' for theta):

cos (t) = L(z)/ L = m_l/ [l (l+1)]^1/2

Of course, it ought to be self-evident that we are obtaining allowed values for the angle, since obviously m_l is going to range from -2 to +2)

Thus, for the problem we've considered (with l = 2) we have the allowed cosines and angles:

m_l = 0 so cos(t) = 0 so t = 90 deg

m_l = 1 so cos (t) = 1/[6]^1/2 so t = 65.9 deg

m_l = -1 so cos(t) = -1/[6]^1/2 so t (theta) = 114.1 deg

m_l = 2 so cos(t) = 2/[6]^1/2 so t = 35. 2 deg

m_1 = -2 so cos(t) = -2/[6]^1/2 = 144.7 deg

2) Find the possible values of L, L(z) and theta for an electron in the 3d state of hydrogen.

This is very simple! For the electron in the 3d state we have: n = 3, l = 2

So: because the configuration fits perfectly with that detailed in (1) all the answers are the same!

3) The orbital angular momentum of the Earth in its orbit around the Sun is:

L = 4.83 x 10^31 kg-m^2/s

Assume it is quantized according to: L = [l (l+1)]^1/2 (h/2pi)and find:

a) the value of l corresponding to the angular momentum, and

b) the fractional change in L as l changes from l to (l +1).

This is not too difficult but one must apply the space quantization principle to the classical orbital angular momentum of Earth. We know:

L = [l(l + 1)]^1/2 (h/2pi) = 4.83 x 10^31 kg-m^2/s

Now (h/2pi) = 1.054 x 10^-34 J-s so that:

L = [[l(l + 1)]^1/2 = (4.83 x 10^31 kg-m^2/s)/ 1.054 x 10^-34 J-s

And:

[l(l + 1)]^1/2 = 4.58 x 10^65

[l(l + 1)] = 2.1 x 10^131

Solve algebraically for l:

l^2 + 1 = 2.1 x10^131

so l~ 4.5 x 10^65

fractional change in L:

Take for the fractional change: (L/L’)

Where L’ = [l+1{(l+1) + 1)}]^1/2 (h/2pi)

L = [l(l + 1)]^1/2 (h/2pi)

Then: L/L’ = [l+1{(l+1) + 1)}]^1/2 / [l(l + 1)]^1/2

L/L’ = [l^2 + l +2]^1/2 / [l(l + 1)]^1/2

L/L’ = 4.49 x 10^65/ 4.58 x 10^65 = 0.98