Quiz 9

Name: Date:Phys224 Spring 2008 Dr. P. Hanlet

Show your work!!! If I can read it, I will give you partialcredit!!! 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

hν =Enhigh−Enlow

L=√

l(l + 1) ~ Lz =ml ~ S =√

s(s + 1) ~ Sz =ms ~

|~µorb|=−e

2m|~L|=µB

√

l(l + 1) µz =−mlµB µs|=−e

m|~S|=2µB

√

s(s + 1) µs,z =−2msµB

U =~µ · ~B Eγ =2µz(Bext + Blocal) λmin =hc

K0

µB =e~

2mms =±

1

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) whichdetermines the energy of the ground state of the atom? Why?The atomic number, Z; i.e. the number of protons in the nucleus of theatom determines the ground state energy of an atom. The reason is thatthe 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 thesame energy level.

3. In your own words, define the Pauli Exclusion Principle. Briefly describe how it contributes to build-ing of atoms.The Pauli Exclusion Principle states that more than one fermion – par-ticles with 1

2-integer spin – cannot exist in the exact same quantum me-

chanical 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 mo-mentum states, and ml =2(l+1) orbital magnetic states, and ms =±1

2spin

states. Once an electron occupies all of the l, ml, and ms states for agiven n, the Pauli Exclusion Principle requires that the next electron filla 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 suchstates.

(c) What are the number of possible values of ms?As ms refers to the intrinsic, or “spin”, angular momentum, it onlyhas values ms =±1

2. 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 differentquantum numbers (recall Pauli’s Exclusion Principle). So, for n=3:

l ml ms #0 0 ±1

22

-1 ±1

2

1 0 ±1

26

1 ±1

2

-2 ±1

2

-1 ±1

2

2 0 ±1

210

1 ±1

2

2 ±1

2

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=3then 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 differ-

ence 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 willgo to the photon; here it is 20% since Kf = 0.8Ki for the electron. Anelectron volt is the energy aquired by a unit charge particle as it is ac-

celerated through a potential difference of 1 V . Here, the initial kineticenergy 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:

Ki =6.4×10−15 J

1.6×10−19 J/eV= 40 keV

Then, the photon energy is:

Eγ = 0.2 · Ki = 0.2 · 6.4×10−15 J = 1.28×10−15 J

And since:

Eγ = hν =hc

λ,

then

λ =hc

Eγ

=

(

6.63×10−34 Js) (

3×108 m/s)

1.28×10−15 J

Finally

λ = 1.55×10−10 m = 155 pm

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