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_____
PHY240-Exam III, Spring 2011
100
Name:____ANSWERS___
PHY240-Modern Physics, Spring 2011 (Exam III)
P1) (18 pts) Quantum Numbers & Degeneracy: An electron is in a state with principal quantum number n=4.
a) List all possible values of the azimuthal quantum number, l. {5 pts}
b) For each of the values in (a), list all possible values of the magnetic quantum number, ml.
{8 pts}
c) Obviously, for every combination of l and ml, there are two spins, ms = + ½, and ms = - ½.
So, what is the degeneracy of the n=4 state? {5 pts}
b)
l=0; ml = 0
a) l=0, 1, 2, 3
l=1; ml = -1, 0, +1
l=2; ml = -2, -1, 0, +1, +2
l=3; ml = -3, -2, -1, 0, +1, +2, +3
c)
l=0; 2 electrons, or sets of quantum numbers (2 spins for each of the ml above)
l=1; 6 electrons or sets of quantum numbers
l=2; 10 electrons or sets of quantum numbers
l=3; 14 electrons or sets of quantum numbers
Giving us a total of 32 different sets of quantum numbers that all have the same n=4.
P2) (15 pts) The Normal Zeeman Effect: When an atom is placed in a magnetic field, there is
a splitting of the energy levels according to values of ml. For example, if l=1, the level is split
like this:
l=1
l=3
ml=-3
ml=-2
ml=-1
ml=0
ml=+1
ml=+2
ml=+3
l=2
ml=-2
ml=-1
ml=0
ml=+1
ml=+2
ml=-1
ml=0
ml=+1
This is referred to as the Normal Zeeman Effect. Suppose an
electron starts in the 5f state, and makes a transition to the 4d
level.
a) What are the possible ml values in the 5f state? What is
the length of L? {5 pts}
b) What are the possible ml values in the 4d state? What is
the length of L? {5 pts}
c) Sketch an energy level diagram (not necessarily to scale)
that shows the splitting in the presence of an external magnetic field. {5 pts}
a) In 5f, 5 means n=5, f means l = 3, so ml = -3, -2, -1, 0, 1, 2, 3
L = 3( 3 + 1) = 12
b) In 4d, 4 means n=4, d means l = 2, so ml = -2, -1, 0, 1, 2
L = 2 ( 2 + 1) = 6
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PHY240-Exam III, Spring 2011
P3) (24 pts) Electrons Transitions and Photon Emissions (and/or absorptions): An electron
in a Hydrogen atom starts in a state which has an excitation energy of 12.09 eV when it suddenly makes a transition to a different state with a binding energy of 3.4 eV.
a) What is ninitial? {3 pts}
b) What is nfinal? {3 pts}
c) Is a photon emitted or absorbed when the electron makes this transition? {3 pts}
d) What is the wavelength of the photon from part (c)? {3 pts}
e) Repeat (a) thru (d) if the electron ends up in a state with a binding energy of 0.54 eV. {12
pts}
(a) Excitation energy is the energy above ground level... which is at -13.6 eV. So:
Einitial = ( −13.6 + 12.09 ) eV = −1.51eV =
−13.6eV
⇒ ninitial = 3
2
ninitial
(b) Binding energy is the energy below vacuum (0 level). So:
E final = −3.4eV =
−13.6eV
⇒ n final = 2
2
n final
(c) Photons are emitted when electrons drop to lower energy states and lose energy. This
electron starts at n=3 and ends at m=2... it loses energy and emits a photon.
(d)
ΔE = E final − E final = −3.4eV − ( −1.51eV ) = −1.89eV
E photon =
hc
λ photon
(e)
E final
⇒ λ photon =
hc
E photon
=
1240eV * nm
= 656nm
1.89eV
Einitial = −1.51eV ⇒ ninitial = 3
−13.6eV
= −0.54eV =
⇒ n final = 5
2
n final
The electron transitions upward, it must absorb a photon’s energy to increase it’s energy...
ΔE = E final − E final = −0.54eV − ( −1.51eV ) = +0.97eV
E photon =
hc
λ photon
⇒ λ photon =
hc
E photon
2
=
1240eV * nm
= 1278nm
0.97eV
PHY240-Exam III, Spring 2011
P4) (5 pts) Hmmm... That’s Strange. (or “What The H#ll is Going On!?!?”): What did
Stern and Gerlack stumble upon as they shot Ag atoms thru a gradient magnetic field in their
attempt to study the orbital angular momentum quantum states of an atom?
They discovered the intrinsic angular momentum we call “spin.”
P5) (20 pts) Atomic Models:
a) Describe J.J. Thomson’s model of the atom. Be sure to mention it’s common
name! {5 pts}
b) Why did Thomson come up with, or devise, this particular model? {5 pts}
c) Describe Ernest Rutherford’s model of the atom. {5 pts}
d) Why did Rutherford come up with, or devise, this particular model? {5 pts}
(a) Thomson’s model was called the PLUM PUDDING MODEL. It consisted of
an amorphous blob of positive charge with electrons embedded in it. Enough
positive stuff to cancel out the charge of whatever number of electrons the
atom has.
(b) Thomson discovered “cathode rays”, or electrons. They seemed to be a constituent part of every different type of atom he looked at, so he had to devise a
model that had internal parts... at least one internal part, the electron.
(c) Rutherford’s model was a nuclear, or planetary model with a very dense, yet
incredibly massive positive nucleus in the very center. Around this there were
electrons in orbits kind of like plants in their orbits... but not necessarily moving in planes like the planets.
(d) Rutherford shot alpha particles (He nuclei) at Au foil and found that most
just flew right through, but some were deflected back, even at angles up to
180°! This meant that most flew through empty space, but some collided head
on with something much more massive then they were and came back.
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PHY240-Exam III, Spring 2011
P6) (18 pts) Most Likely Radii: The wavefunction for the n=3, l=2, ml=0 hydrogen wavefunction is given by:
2
1
1 ⎛ r ⎞ − r 3ao
2
Ψ 320 ( r,θ , φ ) =
e
3cos
θ −1
32 ⎜
⎟
81 6π ( ao ) ⎝ ao ⎠
(
)
(a) Write down just the radially dependent part of the above wavefunction, R(r). {4 pts}
(b) Recalling that P(r) is equal to 4πr2R2(r), write down the radial probability density function, P(r). {4 pts}
(c) At what radial distance(s) are we most likely to find the electron? {10 pts}
2
⎛ r ⎞ − r 3ao
R20 ( r ) = ⎜ ⎟ e
⎝ ao ⎠
4
⎛ r ⎞ −2 r 3ao
2
R20
r
=
( ) ⎜ ⎟ e
⎝ ao ⎠
4
⎛ r⎞
4π
P ( r ) = 4π r 2 ⎜ ⎟ e−2 r 3ao = 4 r 6 e−2 r 3ao
ao
⎝ ao ⎠
dP ( r ) 4π d 6 −2 r 3ao
= 4
r e
dr
ao dr
⎞
dP ( r ) 4π ⎛ 5 −2 r 3ao
6 ⎛ −2 ⎞ −2 r 3ao
= 4 ⎜ 6r e
+r ⎜
e
⎟ =0
dr
ao ⎝
⎝ 3ao ⎟⎠
⎠
⎛
⎛ −2 ⎞ ⎞
⎛ 2 ⎞
6
+
r
=
0
⇒
r
=6
⎜
⎟
⎜
⎟
⎜
⎟
⎝ 3ao ⎠ ⎠
⎝ 3ao ⎠
⎝
(
r = 9ao = 32 ao
)
Note that this is just n2ao, as the Bohr Model predicted for the radius of an orbit! Woo-hoo!
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