At the end of the previous blog I showed the solution for the electron’s intrinsic angular momentum for m_s = ½. The diagram appended at left shows this graphically. Let’s now look at a few simple problem in QM.

First, a brief refresher: Recall the probability of finding a particle in some interval (a,b) is:

First, a brief refresher: Recall the probability of finding a particle in some interval (a,b) is:

P(a,b) = INT(b,a) [U]^2dx

For a one dimensional system where U(x) is the wave function and INT(b,a) denotes the integral from a to b.

Problem: Let U(x) = [2/L]^1/2 sin (2 pix/L)

Find the probability of a particle governed by the wave function U(x) to be found between x = L/4 and x = 0.

Here: b = L/4 and the limit a = 0

So; P(a,b) = INT(L/4, 0) ([2/L]^1/2)^2 sin^2(2 pi x/L)dx

We can simplify the above using a trigonometric identity:

Angle Q = 2 pi x/ L so: sin^2Q = ½ (1 – cos 2Q)

Then:

P(a,b) = 2/L INT(L/4, 0) {½ (1 – cos (4pix/L))}dx

P(a,b) = 2/L [{½INT(L/4, 0)dx - INT(L/4, 0) cos(4pix/L)}dx

P(a,b) = [x/L – 1/2pi sin (4pix/L)] (with x = L/4 and x = 0, then subtract)

P(a,b) = ¼ – 0 = 0.25

Problem 2. (Readers may find it useful to review this material: http://brane-space.blogspot.com/2010/06/more-on-quantum-numbers.html

Write out the electron configuration for oxygen (O16), then write out the values for the set of quantum numbers : n, l, m_l, m_s for each of the electrons in O16

Finding the electron configuration means one must use the Pauli Exclusion Principle to make sure the electrons are distributed so that no two electrons have the same set. This means we need to have 2 in the 1s shell, 2 in the 2s shell, and 4 in the 2p shell, so:

1s(2) 2s(2) 2p(4)

For the2 electrons in 1s shell: n = 1, l = 0, m_l= 0 and m_s = ½ (and (- ½), for second)

For the 2 electrons in the 2s shell: n=2, l =0, m_l = 0 and m_s = ½ (and (- ½), for second)

For the 4 electrons in the 2p shell:

Since l=1 corresponds to p: n=2, l=1, m_l= +1, m_s = + ½

And: n=2, l=1, m_l= -1, m_s = -½

And: n=2, l=1, m_l= 0, m_s = +½

Finally: n=2, l=1, m_l= 0, m_s = -½

More problems to come next time!

For a one dimensional system where U(x) is the wave function and INT(b,a) denotes the integral from a to b.

Problem: Let U(x) = [2/L]^1/2 sin (2 pix/L)

Find the probability of a particle governed by the wave function U(x) to be found between x = L/4 and x = 0.

Here: b = L/4 and the limit a = 0

So; P(a,b) = INT(L/4, 0) ([2/L]^1/2)^2 sin^2(2 pi x/L)dx

We can simplify the above using a trigonometric identity:

Angle Q = 2 pi x/ L so: sin^2Q = ½ (1 – cos 2Q)

Then:

P(a,b) = 2/L INT(L/4, 0) {½ (1 – cos (4pix/L))}dx

P(a,b) = 2/L [{½INT(L/4, 0)dx - INT(L/4, 0) cos(4pix/L)}dx

P(a,b) = [x/L – 1/2pi sin (4pix/L)] (with x = L/4 and x = 0, then subtract)

P(a,b) = ¼ – 0 = 0.25

Problem 2. (Readers may find it useful to review this material: http://brane-space.blogspot.com/2010/06/more-on-quantum-numbers.html

Write out the electron configuration for oxygen (O16), then write out the values for the set of quantum numbers : n, l, m_l, m_s for each of the electrons in O16

Finding the electron configuration means one must use the Pauli Exclusion Principle to make sure the electrons are distributed so that no two electrons have the same set. This means we need to have 2 in the 1s shell, 2 in the 2s shell, and 4 in the 2p shell, so:

1s(2) 2s(2) 2p(4)

For the2 electrons in 1s shell: n = 1, l = 0, m_l= 0 and m_s = ½ (and (- ½), for second)

For the 2 electrons in the 2s shell: n=2, l =0, m_l = 0 and m_s = ½ (and (- ½), for second)

For the 4 electrons in the 2p shell:

Since l=1 corresponds to p: n=2, l=1, m_l= +1, m_s = + ½

And: n=2, l=1, m_l= -1, m_s = -½

And: n=2, l=1, m_l= 0, m_s = +½

Finally: n=2, l=1, m_l= 0, m_s = -½

More problems to come next time!

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