1

Physics 9HE-Modern Physics Sample Final Exam (100 points total)

---------------------------------------------------------------------------------------------- Miscellaneous data: c = 3.00 x 108 m/s e = 1.60 x 10-19 C 1 eV = 1.60 x 10-19 J 1 Å = 10-10 m

MSun = 2 x 1030 kg MEarth = 5.98 x 1024 kg rEarth = 6.38 x 106 m

me = 9.1094 x 10-31 kg = 0.5110 MeV/c2 mp = 1.6726 x 10-27 kg = 938.27 MeV/c2

mn = 1.6749 x 10-27 kg = 939.57 MeV/c2 1 u = 1.6605 x 10-27 kg = 931.49 MeV/c2

m(1H) = 1.0078 u G = 6.67 x 10-11 Nt-m2/kg2 g = 9.81 m/s2 σ = 5.67 x 10-8 W-m-2-K-4

h = 6.63 x 10-34 J-s h = h/2π = 1.05 x 10-34 J-s kB = 1.38 x 10-23 J-K-1 a0 = 0.529 Å

N(t) = Noexp(-t/τO) = Noexp(-0.693t/τ1/2) kC ≡ 1/(4πεo) = 8.98 x 109 N-m2-C-2 RH = 1.09678 x 107 m-1 2

2 2 1/ 2 2 1/ 21 1 1 0.5

(1 ) (1 )if v c

v cγ β

β= = ≅ +

− − β = v/c = [(γ2-1)/γ2]1/2

T= γTo L = Lo/γ 1/ 2

0 01/ 2(1 ) [1 ] 1(1 )

forβν ν ν β ββ

±= ≅ ± <<

m ν = c/λ ν (lect.) = f(book)

( ) 2' ' ' ' vx x vt y y z z t t xc

γ γ ⎡ ⎤⎛ ⎞= − = = = − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

' ' '

2 2 21 1 1

yx zx y z

x x x

uu v uu u uv v vu u uc c c

γ γ

−= = =

⎛ ⎞ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

x2 + y2 + z2 –c2t2 = invariant

2 2 2 2 2 2 4p mu E mc K mc E p c m cγ γ= = = + = +r r

2 2 21 2

1 1 2( 1) SchT GM gH GME pc h if R

T r rc c cν ν λ ννν ν λ ν

⎡ ⎤Δ Δ Δ Δ Δ= = = = − − ≈ − ≈ − << =⎢ ⎥

⎣ ⎦

2tan

e

e Vm dB

θ=

l λmaxT = 2.898 x 10-3 m-K

2

52 1( , )

1hckT

c hI Te λ

πλλ

= ⋅−

R(T) = εσT4

2 2 221 2

2 2 4( )16 4 sin ( / 2)i

o

N nt Z ZeNr K

θπε θ

⎛ ⎞= ⎜ ⎟

⎝ ⎠

21 2 cot( / 2)

8 o

Z Z ebK

θπε

= 2

1 2min 4 o

Z Z erKπε

= 2( )f nt nt bθ σ π≥ = =

12h' (1 cos ) 2.43 10 (1 cos ) ( in m )mc

Δλ λ λ θ θ−= − = − = × − hν = Kmax + φ minhceV

λ = nλ = 2dsinθ 2

1 2coul C radial2

q q mvF k Frr

= = 22 2 2 2

20n n 02 2 2

0 0 e

4Z e 13.6Z nE (eV ) r n aZ8 a n n m Ze

πεπε

= − = − = =h

20

0 24 0.529

e

am eπε

= =h Å n H e e

e n eu

n Mv Z R mm r M mn n

μλ

⎡ ⎤ ⎡ ⎤= = − =⎢ ⎥ ⎢ ⎥+⎣ ⎦⎣ ⎦

22 2

1 1 1l

h

λ = h/p p = hk ΔxΔpx ≥ h/2 ΔEΔt ≥ h/2 ωphv = /k ωgr v = d /dk

± − −⎡ ⎤ ⎡ ⎤= ± = + = −⎣ ⎦ ⎣ ⎦ix ix ix ix ix1 1e cos x i sin x cos x e e sin x e e

2 2i

sin 2t = 2 sin t cos t cos 2t = cos2 t – sin2 t = 2 cos2 t – 1 = 1 – 2 sin2 t

2

π π πψ

π πψ ψ ψ

∞ ∞

= =

= + + =

= = =

∑ ∑

∫ ∫l l

l l l

ll l l l

0n n n

n 1 n 1

0 n n0 0

a 2 n 2 n 2 n( x ) a cos( x ) b sin( x ), with k , and2

2 2 n 2 2 na aver . of over , a ( x')cos( x')dx', b ( x')sin( x')dx'

ikx ikx'1 1( x ) c(k )e dk, with c(k ) ( x')e dx'2 2

ψ ψπ π

∞∞ −

−∞−∞

= =∫ ∫

∂ΨΨ∂

= hH it = +ˆ ˆH K V ωΨ ψ ψ

− −= =h

iE ti te e ψ ψ=H E

xp ix∂∂

= − h 2 2

22x 2K A A A

2m x∂ Δ∂

= − = − ∞h

ψ ψ=a aA a ˆA * A dxψ ψ< >= ∫

ikx x

2 22m(E V ) 2m(V E )k e sin kx,cos kx e κψ ψ κ α ψ± ±− −

= ∝ ∝ = = ∝h h

1 / 2 2 2 2

n n 22 n x nsin EL L 2mL

π πψ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

h

2

2 1x / 2

n n n rot2m 1 ( 1) mRH ( x )e E n ; E I

m 2 2I 2α κ κψ α ω ω− +⎛ ⎞= = = = + = =⎜ ⎟

⎝ ⎠

h l lh

h

1 12 2 2 22 L0 0

0 0 0 0

V sin ( kL ) V sinh ( L ) E ET 1 T 1 16 1 e ( when L 1 )4E ( E V ) 4E (V E ) V V

κκκ

− −

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + = + ≈ − >>⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3 / 2e 2KL

FE STM

4 2mT exp I edV3e

dx

φ −

⎡ ⎤⎢ ⎥

≈ − ∝⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

h nuc

coll 15

ZR ( m )0.0993 MeVf T T exp 4 Z 8E ( MeV ) 7.3 10α α

α

λ π −

⎡ ⎤= = − +⎢ ⎥×⎣ ⎦

2

2 2n m n m m n m n n

B s B

24 2B N

e p

x r sin cos y r sin sin z r cos dV r dr sin d d

( r , , ) R ( r ) ( ) ( ) R ( r )Y ( , ) P ( r ) r R ( r )

e L e siA E B L s 22m m

e e9.274 10 J / Tesla 5.058 102m 2m

θ φ θ φ θ θ θ φ

ψ θ φ Θ θ Φ φ θ φ

μ μ μ μ μ μ

μ μ− −

= = = =

= = =

= = − = − = − = − = −

≡ = × ≡ = ×

l l l ll l l l l l l

l

r rr r rr r r

h h

h h 7 J / Tesla

MOj Ai , j Ai

Atoms AOrbitals i

( r ) c ( r )ψ ϕ= ∑r r

[ ]

2 2A 1 A 2 2 / 3Z n Z V A S

0 nuc15 1 / 3 2

nuc initial final

3 Z( Z 1)e (N Z )B( X ) [Nm Zm( H ) M( X )]c a A a A a5 4 R A

R (1.2 10 m )A Q M M c

δπε

−

− −= + − ≈ − − − +

≈ × = −

---Tear off this sheet and begin exam---

1

Physics 9HE-Modern Physics Sample Final Examination

(100 points total)

Name (printed)_____SAMPLE ANSWERS_____ Name (signature)_____________________________________ Student ID No.____________________________________

[1.1] (10 points) The free neutron at rest decays into a proton and an electron with an exponential lifetime of τ0 = 920 s. Now consider a beam of neutrons with a kinetic energy of 100 MeV travelling along the +x direction relative to an observer in the laboratory. (a) At what speed will the neutrons in the beam be moving?

(b) What would the observer in the laboratory measure for the lifetime of the neutrons in the beam?

2

[1.2] (25 points) Answer the following questions with brief statements or calculations (a) State three types of experimental observations that challenged classical physics around 1900.

Any three of:

Blackbody radiation Electromagnetism and the origin of magnetic fields, which did not transform

from one coordinate system to another Line spectra of hydrogen and other atoms The photoelectric effect Compton scattering (b) State three experimentally verified consequences of Special Relativity. Any three of: Time dilation, e.g. in planes circling earth Length contraction, e.g. in muon viewing mountain Relativistic momentum: effective mass increase of moving object as viewed in fixed frame Relativistic energy: E = mc2

Special version of Doppler effect B fields as a relativistic manifestation of E fields (c) So-called L x-rays are emitted from a copper atom in which an initial 2p vacancy is created. What transitions from the n = 3 shell are permitted in generating these x-rays and why? Dipole selection rules are: Δl = ±1 and Δml = 0, ±1, so the allowed transitions are from 3d with Δl = 1-2 = -1 and 3s with Δl = 1-0 = +1, and this would get full credit. In more detail using the other selection rule, we would have:

_______ _______ _______ _______ _______

_______

_______ _______ _______

3d: l = 2, ml = -2 -1 0 +1 +2

3s: l = 0, ml = 0

2p: l = 1, ml = -1 0 +1

_______ _______ _______ _______ _______

_______

_______ _______ _______

3d: l = 2, ml = -2 -1 0 +1 +2

3s: l = 0, ml = 0

2p: l = 1, ml = -1 0 +1

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(d) A red laser is used to produce a perfect sinusoidal traveling light wave of frequency νo = 1014 Hz. However, by special means, the wave is abruptly cut off at each end so that it has a finite length in time of 10-13 s. If this finite wave is now described in a Fourier representation, what would be the approximate range of frequencies Δν involved? (d) Define degeneracy and give one example of it from the systems we have studied.

(e) The position of an electron along the x direction is measured with an uncertainty of 1 Å. What can you say about the uncertainty in its x momentum? What, if anything, can you say about the uncertainty in its y momentum? (f) In the oxygen molecule pointing along the x direction, one of the bonding wave functions can be made up of oxygen 2px functions on the two atoms. Indicate the

zz

4

equation for this wave function, and sketch it, using a 3D contour of equal probability density.

x xMolecular O2 p on atom1 O2 p on atom 2Orbital

(Simple s ign OK too )

ψ φ φ∝ −

+

- -+ +- -+ + x

5

[1.3] (15 points) Consider a hydrogenic atom consisting of a single electron around a uranium nucleus (Z = 92). (a) What will be the energy in eV and the radius in Å of the first Bohr orbit for this atom?

(b) Balance coulombic force and acceleration and use the result of part (a) to derive the speed of the electron in the first Bohr orbit around this atom, using a non-relativistic approach. Based on your answer, is a non-relativistic model adequate? (c) Now consider this hydrogenic atom initially in an n = 10 excited state, from which it deexcites to the n = 1 state by emitting electromagnetic radiation. What would be the weight change of the atom due to emitting this radiation, including its sign (increase or decrease)?

2 22 1910 1

2 2 8 2

35 31

1 1{ }E EE 1 10m (13.6eV )(92 ) (1.602x10 J / eV )c c (3x10 )

20,284x10 kg 2.03x10 kg, about 1 / 5 the mass of an electron!

Energy lost , so mass decreases : m negative

−

− −

−−= = =

= =

=

ΔΔ

Δ

cc

6

[1.4] (10 points) Consider an electron trapped inside the potential well shown below, at an energy of 200 eV, at the position indicated: (a) Qualitatively sketch on the diagram the form of the wavefunction in each of the regions 1-4 indicated, being careful to show the relative wavelengths and the relative amplitudes in each region. (That is, in which region of the box will the particle be most likely to be found?)

From (c) below, the wavelength is pretty small compared to this well’s dimensions, so there are something like 50/0.868 = 57 cycles in each of regions 2 and 3. Too hard to draw this, but I have indicated relative heights of ψ such that |ψ|2 is inversely proportional to the particle velocity in the region (a correspondence principle type of argument), which is in turn proportional to the square root of kinetic energy. So |ψ2|2/|ψ3|2 ≈ [KE3/KE2]1/2 = [100/200]1/2 = 0.707 and finally ψ2/ψ3 ≈ 0.840. Particle is more likely to be found in region 3. Full credit here was simply showing a smaller wavelength and amplitude in region 2.

(b) Indicate the classical turning points appropriate to this state on the diagram as well. See diagram. Two of them.

∞…∞

0 Å 50 Å 100Å x

e-

1 2 3 4

• •Classical turning points

sin k2x

Zero here

sin k3x, cos k3x

Exponential decay:exp(-κ4x)

Higher amplitude andLonger wavelength in 3

∞…∞

0 Å 50 Å 100Å x

e-

1 2 3 4

∞…∞

0 Å 50 Å 100Å x

e-

∞…∞

0 Å 50 Å 100Å x

e-

1 2 3 4

• •Classical turning points

sin k2x

Zero here

sin k3x, cos k3x

Exponential decay:exp(-κ4x)

Higher amplitude andLonger wavelength in 3

7

(c) For the energy values shown on the diagram, what is the wavelength of the particle in region 2? In general, λ = h/p, and in non-relativistic limit energy of particle in region 2 = E2 = KE = kinetic energy = p2/2m, so p = sqrt[2mE] and finally λ = h/sqrt[2mE]. With E = 200 eV = 200(1.60 x 10-19 J/eV) = 3.20 x 10-17J, we thus have finally, in SI units: λ = h/sqrt[2mE] = [6.63 x 10-34J-s]/sqrt[2(9.11 x 10-31kg)(3.20 x 10-17J) = [6.63 x 10-34]/7.63 x 10-24 = 0.868 x 10-10 m = 0.868 Å [1.5] (15 Points) Consider an electron incident from the left side of the infinitely long potential step at x = 0 shown below. The particle energy E = 10 eV and the step has a height of Vo = 7 eV. (a) What is the relevant time-independent Schroedinger equation to the left of the step (region I) and to the right of the step (region II)? You need not solve these equations.

x

e-

E = 10 eVV0 = 7 eV

I II

x

e-

E = 10 eVV0 = 7 eV

x

e-

E = 10 eVV0 = 7 eV

I II

8

(b) What would be the time-dependent form of the wave function for this problem in both regions, including both transmitted and reflected components and taking advantage of the known general solutions to this type of problem?

9

(d) Now set up the boundary conditions needed to solve this problem, but you need not go beyond this. [1.6] (15 Points) Two of the 3d hydrogen-atom wavefunctions are given by:

o

2

r3a2 2 2i

3 ,2 , 2 3d( r , , ) Cr e sin e φψ θ φ ψ θ±

−±

± = = (a) Indicate how the probability density associated with these wavefunctions would be calculated, and how the normalization constant C would be determined from this density, going as far as you can without evaluating any non-trivial integrals.

(b) Sketch and describe in words in an unambiguous way the three-dimensional probability density of these two hydrogenic wave functions, using either a 3D contour of equal probability or a plot in which greater darkness means a higher probability of finding the electron at a given position.

OK to this lineOK to this line

10

(c) Write down three eigenfunction relations satisfied by these wave functions, including a specification of the eigenvalues involved and the physical meaning of the eigenvalue.

(d) Write down the time-dependent form of these two wave functions.

Just add the usual complex exponential as:

6o

2

r3a2 2 2i i t

3 ,2 , 2 3d

2

3 2

( r , , ) Cr e sin e e

1 1.51 eVwhere E / (13.6eV )3

φ ωψ θ φ ψ θ

ω

±

−± −

± = =

= = − = −hh h

[1.7] (10 Points) Consider the nuclide 49

24 xCr of chromium, with atomic mass of 48.951341 u. (a) What is x here, and what is the binding energy of this nuclide per nucleon? (b) This nuclide decays by positron emission to form a nuclide of vanadium = V. Write down the overall reaction, including the new nuclide that would be formed. Positron emission reduces the atomic no. Z by one, and overall reaction is: 49 49

24 25 23 26 eCr V e ν+→ + +

11

OK without neutrino stated. (c) Which one of the four fundamental interactions is responsible the force between quarks and what is the particle mediating this force?

The strong interaction is responsible for quark interaction and the mediating particle is the gluon.

---End of examination---