Download Quiz 9

Quiz 9
Name:
Phys224
Date:
Spring 2008
Dr. P. Hanlet
Show your work!!! If I can read it, I will give you partial
credit!!! Correct answers without work will NOT get full credit.
Potentially useful equations: (recall that I use ν = f for frequency)
c = 3×108 m/s
|qe | = |qp | = 1.6×10−19 C
h = 6.63×10−34 J·s = 4.14×10−15 eV·s
me = 9.11×10−31 kg
p
p
hν = Enhigh−Enlow
L = l(l + 1) ~
L z = ml ~
S z = ms ~
S = s(s + 1) ~
p
p
e ~
e ~
|~µorb| = −
µz = −ml µB
µs | = − |S|
µs,z = −2ms µB
|L| = µB l(l + 1)
= 2µB s(s + 1)
2m
m
hc
e~
1
~
U =µ
~ ·B
Eγ = 2µz (Bext + Blocal )
λmin =
µB =
ms = ±
K0
2m
2
l = 0, 1, 2, . . . , (n − 1)
ml = −l, −(l−1), . . . , −2, −1, 0, 1, 2, . . . , (l−1), l
Concept (3 points)
1. For electrons bound in atoms, what is the principle factor (other than mass and constants) which
determines the energy of the ground state of the atom? Why?
The atomic number, Z; i.e. the number of protons in the nucleus of the
atom determines the ground state energy of an atom. The reason is that
the potential experienced by an electron depends on the total charge,
hence Z, of the nucleus.
2. In your own words, what are degenerate states.
Degenerate states are different states with the same energy; i.e. in the
same energy level.
3. In your own words, define the Pauli Exclusion Principle. Briefly describe how it contributes to building of atoms.
The Pauli Exclusion Principle states that more than one fermion – particles with 21 -integer spin – cannot exist in the exact same quantum mechanical state; i.e. at least one quantum number must differ. As a result,
for any energy level, n, there are l = 0, 1, . . . , (n − 1) orbital angular momentum states, and ml = 2(l+1) orbital magnetic states, and ms = ± 21 spin
states. Once an electron occupies all of the l, ml , and ms states for a
given n, the Pauli Exclusion Principle requires that the next electron fill
a higher energy state, n+1.
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Problem (5 points)
4. An electron is in a state with n = 3.
(a) What are the possible values of l?
The possible values of l are:
l = 0, 1, 2, . . . , (n−1)
Hence, if n = 3, then here:
l = 0, 1, 2
(b) What are the number of possible values of ml if l has its maximum value?
The possible values of ml are:
ml = −l, −(l−1), −(l−2), . . . , −2, −1, 0, 1, 2, . . . , (l−2), (l−1), l
Hence, if l = 2, then ml = −2, −1, 0, 1, 2, such that there are 5 such
states.
(c) What are the number of possible values of ms ?
As ms refers to the intrinsic, or “spin”, angular momentum, it only
has values ms = ± 12 . Hence there are only 2 possible values of ms .
(d) What are the number of states in the n = 3 shell?
The number of states is the total number of combinations of different
quantum numbers (recall Pauli’s Exclusion Principle). So, for n = 3:
l
ml
ms
#
0
0
-1
0
1
-2
-1
0
1
2
± 21
± 12
± 21
± 12
± 12
± 12
± 21
± 12
± 12
2
1
2
6
10
Hence, there are 18 possible states in the n = 3 energy level.
(e) What are the number of subshells in the n = 3 shell?
The subshells for a given n level are given by the different values of l.
There are always the same number of l values as n. Here, since n = 3
then there are 3 different subshells.
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5. X-rays are produced in an x-ray tube by electrons accelerated through an electric potential difference of 40 kV . Let Ki be the kinetic energy of the electron at the end of the acceleration. After
the electron collides with the target nucleus (assume nucleus remains stationary), the electron has
kinetic energy Kf = 0.8Ki . What is the wavelength of the associated photon?
Conservation of energy dictates that the energy lost by the electron will
go to the photon; here it is 20% since Kf = 0.8Ki for the electron. An
electron volt is the energy aquired by a unit charge particle as it is accelerated through a potential difference of 1 V . Here, the initial kinetic
energy of the electron is:
Ki = qV = 1.6×10−19 C 4.0×104 J/C = 6.4×10−15 J
ASIDE:
To convert this to electron-volts:
6.4×10−15 J
Ki =
= 40 keV
1.6×10−19 J/eV
Then, the photon energy is:
Eγ = 0.2 · Ki = 0.2 · 6.4×10−15 J = 1.28×10−15 J
And since:
hc
,
λ
6.63×10−34 Js 3×108 m/s
hc
λ=
=
Eγ
1.28×10−15 J
Eγ = hν =
then
Finally
λ = 1.55×10−10 m = 155 pm
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