Physics 249 Final Exam
Dec 21st 2012
Constants
Proton mass: 1.6726x10-27 kg or 938.27 MeV/c2
Electron mass: 9.1094x10-31 kg or 0.51100 MeV/c2
Electron charge: e=1.60x10-19 C
ke = 8.99x109 Nm2/C2
𝑘2 𝑒 2 = 1.44𝑒𝑉𝑛𝑚
ℏ = 1.0546x10-34 J*s, 6.5821x10-16 eV*s
c = 3.00x108 m/s
ℎ𝑐 = 1240𝑒𝑉𝑛𝑚
ℏc = 197.3eVnm
The mass energy of 1 atomic mass unit, u is 931.5 MeV
Integrals
1) a) What is the minimum energy(in eV) and wavelength(in nm) of a photon needed to ionize the n=3 orbit
of He+.
b) If you use a photon of wavelength 100nm to ionize the electron in this system what is the stopping
potential (in V) for the ionized electron.
(15 points)
a)
2
𝜇 𝑘𝑍𝑒 2
𝑍2
22
𝐸𝑛,𝑍 = − 2 (
) = 13.6 2 𝑒𝑉 = 13.6 2 𝑒𝑉 = 6.04𝑒𝑉
2𝑛
ℏ
𝑛
3
ℎ𝑐 1240𝑒𝑉𝑛𝑚
𝐸 = ℏ𝜔 =
=
𝜆
𝜆
1240𝑒𝑉𝑛𝑚
𝜆=
= 205𝑛𝑚
𝐸
b)
𝑒𝑉0 = ℏ𝜔 − 𝜙 =
ℎ𝑐
1240𝑒𝑉𝑛𝑚
−𝜙 =
− 6.04𝑒𝑉 = 6.36𝑒𝑉
𝜆
100
𝑉0 = 6.36 𝑉
2) A gold foil is used in a Rutherford experiment to scatter alpha particles with energies 10 MeV. What
fraction of the particles are backscattered to angles within 5 degrees of the incoming direction.
For the foil. Z=79, thickness = 1um, n = 5.90e28 atoms/m 3
(15 points)
The number scattered to an angle greater than theta is given by:
𝑁 = 𝜋𝑏 2 𝐼𝑛𝑡𝐴
while the total number of incident particles is simply.
𝑁 = 𝐼𝐴
so
𝑓 = 𝜋𝑏 2 𝑛𝑡
where b is given by
𝑏=
𝑘𝑞𝑄
𝜃 𝑘𝑞𝑄
𝜃 2 ∗ 79 ∗ 1.44𝑒𝑉𝑛𝑚
175
𝑐𝑜𝑡 =
𝑐𝑜𝑡 =
cot (
) = 4.97𝑥10−7 𝑛𝑚 = 4.97𝑥10−16 𝑚
2
𝑚𝑣
2 2𝐾𝐸
2
2 ∗ 1𝑒7
2
𝑓 = 𝜋(4.97𝑒 − 16𝑚)2 (5.90e28)(1𝑒 − 6) = 4.57𝑥10−8
3) a) For the Harmonic oscillator wave functions n=0, 1, 2 prove for all cases the equation.
+∞
𝜓𝑛∗ 𝑥𝜓𝑚 = 0
∫
−∞
unless 𝑛 = 𝑚 ± 1
b) Interpret what this result means for transitions between different energy states of the harmonic oscillator
𝜓0 = 𝐴0 𝑒 −𝑦
2 /2
𝜓1 = 𝐴1 𝑦𝑒 −𝑦
2 /2
𝜓2 = 𝐴2 (2𝑦 2 − 1)𝑒 −𝑦
𝑦 = √𝛼𝑥,
𝛼=
2 /2
ℏ𝜔
2
(25 points)
a)
for n=m
+∞
+∞
𝜓𝑛∗ 𝑥𝜓𝑛 ∝ ∫
∫
−∞
𝜓𝑛∗ 𝑦𝜓𝑛 = 0
−∞
you can think of this as the expectation value of x which is zero or as manifestly odd function which is zero
when integrated over a symmetric region.
+∞
+∞
𝜓0∗ 𝑥𝜓1 = ∫
∫
−∞
𝐴0 𝑒 −𝑦
2 /2
𝑥𝐴1 𝑦𝑒 −𝑦
2 /2
−∞
finite since this is an even function
+∞
∫
−∞
+∞
𝜓1∗ 𝑥𝜓2 = ∫
−∞
𝐴1 𝑦𝑒 −𝑦
2 /2
+∞
∝∫
2
−∞
𝑥𝐴2 (2𝑦 2 − 1)𝑒 −𝑦
can be finite since it is an even function.
𝑦 2 𝑒 −𝑦
2 /2
+∞
∝∫
−∞
𝑦 2 (2𝑦 2 − 1) 𝑒 −𝑦
2
+∞
2
𝑦 2 (2𝑦 2 − 1) 𝑒 −𝑦 = 4 ∗
∫
−∞
+∞
+∞
𝜓0∗ 𝑥𝜓2 = ∫
∫
−∞
𝐴0 𝑒 −𝑦
720 ∗ √𝜋 √𝜋
−
≠0
8
4
2 /2
𝑥𝐴2 (2𝑦 2 − 1)𝑒 −𝑦
−∞
2 /2
+∞
∝∫
2
𝑦(2𝑦 2 − 1) 𝑒 −𝑦 = 0
−∞
zero since this is an odd function integrated over a symmetric region.
b) This defines a selection rule for what transitions are allowed between different harmonic oscillator
energy states.
4) Use the Heisenberg uncertainty principle to estimate the size of the hydrogen atom. Calculate the
expectation value of r to determine the average radius of the ground state and compare this to the estimate
from the uncertainty principle.
1 2 −𝑟/𝑎
0
𝑒
4𝜋 √𝑎03
(20 points)
𝜓=√
Using:
Δ𝑥Δ𝑝 >
ℏ
2
At the minimum energy which should be similar to the ground state energy.
𝑝~
ℏ
2𝑟
The total energy of the electron is the sum of it’s kinetic and potential energies.
𝐸=
𝑝 2 𝑘𝑒 𝑒 2
ℏ2
𝑘𝑒 𝑒 2
−
=
−
2𝑚
𝑟
8𝑚𝑟 2
𝑟
to find the minimum
𝑑𝐸 −2ℏ2 𝑘𝑒 𝑒 2
=
+ 2 =0
𝑑𝑟 8𝑚𝑟 3
𝑟
2ℏ2
𝑘𝑒 𝑒 2
=
8𝑚𝑟 3
𝑟2
𝑟=
ℏ2
1
= 𝑎0
2
4𝑚𝑘𝑒 𝑒
4
for the expectation value
𝜋,2𝜋
⟨𝑟⟩ = ∬
0
∞
∫ (√
0
1 2 −𝑟/𝑎
1 2 −𝑟/𝑎
0 ) 𝑟 (√
0 ) 𝑟 2 𝑑𝑟𝑠𝑖𝑛𝜃𝑑𝜃𝑑𝜙
𝑒
𝑒
4𝜋 √𝑎03
4𝜋 √𝑎03
⟨𝑟⟩ =
4 ∞ 3 −2𝑟/𝑎
4 3! 𝑎04 3
0 𝑑𝑟 =
∫
𝑟
𝑒
= 𝑎0
2
𝑎03 0
𝑎03 16
The answer is off by a factor of 6 which is within an order of magnitude. This is not the harmonic
oscillator so we don’t expect the Heisenberg uncertainty estimates to be exact.
19
5) Consider 10
𝑁𝑒 and 199𝐹
a) What orbitals are filled for the protons and neutrons in these two nuclei? List how many neutrons or
protons are in each energy level. Use the notation that lists energy levels by primary quantum number,
orbital shell type and total angular momentum, rather than shell number. (Hint – all of the levels that are
filled have n=1)
b) Which is the unstable isotope and why.
c) Write down the decay mechanism that converts the unstable to the stable isotope.
d) Calculate the nuclear radius for these isotopes given
931.5𝑀𝑒𝑉
∆𝑈 = (19.001880 − 18.998404)𝑢 ∗ (
)
𝑢
(15 points)
a) A complete 1st shell for both protons and neutrons in both atoms 2 1s, 4 p3/2 and 2 p1/2
In the second shell
For Ne. protons: 2 1d5/2, 1 1d5/2 neutrons
For F, protons: 1 1d5/2, 2 1d5/2 neutrons
b) 19 Ne is the unstable isotope because the extra proton adds proton coulomb repulsion to the potential
energy which will be lowered if 19 Ne decays to 19 F
c)
19
10𝑁 𝑒
→
19
9𝐹
+ 𝑒 + + 𝜈𝑒
d)
∆𝑈 =
3 𝑘𝑒 2 2
(𝑍 − (𝑍 − 1)2 )
5 𝑅
931.5𝑀𝑒𝑉
3 1.44𝑥10−6 𝑀𝑒𝑉𝑛𝑚
(19.001880 − 18.998404)𝑢 ∗ (
(102 − (10 − 1)2 )
)=
𝑢
5
𝑅
R=5.07e-15m
6) Annihilating electrons and positrons to photons followed by pair production has been one of High
Energy physics basic tools for producing and studying new particles.
Write Feynman diagram for the following process
a) One diagram for 𝑒 + 𝑒 − → 𝜇 + 𝜇 −
In some cases more diagrams are possible and you have to add the applitudes of both interactions.
b) Write two diagrams for 𝑒 + 𝑒 − → 𝑒 + 𝑒 −
c) Give one example of a particle antiparticle pair that cannot be produced this way.
(10 points)
a) Annihilation diagram
b) Annihilation diagram and scattering diagram
c) Neutrino anti-neutrino pairs.