Past Papers Solutions Output

8/10/2019 Past Papers Solutions Output

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Solutions


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Equations for Light

E = hc =

DeBroglie Relation

h= p

Operators

H = p2

2m+ V

H(harmonic oscillator) = p2

2m

+1

2

kx2

H(rotation) = J2

2II = mr2

H(hydrogenic) = p2

2m

Z e2

4o r[A, B] = A B B A

Aave =

Add

= m1m2m1+ m2

Eigenvalues and Eigenfunctions

particle in a box: En = n2h2

8mL2, n= 1, 2, 3, 4, . . . n(x) =

2

Lsin(nx/L)

harmonic oscillator: En = (n + 1

2)ho, n= 0, 1, 2, 3, . . . n(x) =NnHn(y)ey

2/2

o = 12

km

rigid rotor: Ej = j(j+ 1)h2

2I,

j = 0, 1, 2, 3, . . .m= j, . . . , j

Yj, m(, ) = j, m()ei m

J2 Yj, m = j(j+ 1)h2Yj, m

JzYj, m = mhYj, m

1-electron atom: En = Z2

n2 h2

2mea2o ,

n= 1, 2, 3, . . .

j= 0, . . . , n 1m= j, . . . , j

n j m(r,,) = Rn j(r)Yj, m

n j m s ms =

n j m or

n j m

Rn j(r) =rj (polynomial in r of order [n-j-1])

P.T.

Heisenberg Uncertainty Principle

pq h

2

AB [A, B]ave

2

Radial Distribution Function

P(r) =r2RR

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Fundamental constants and conversion factors

h = 6.626 1034Js

c = 3 108ms1

R = 8.31JK1mol1

me = 9.11 1031kg

ao = 0.529 1010m

1J = 1kg m2/s2

Complex numbers

eikx = cos(kx) + i sin(kx)

Spherical Polar Coordinates

dV =r2 dr sin d d

Integrals

sin(nx/L)dx = (L/n)cos(nx/L) sin2(nx/L)dx = 0.5x (L/4n) sin(2nx/L)

0

rn ea r dr = n!

an+1

.

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1. [5 marks] |(x)|2 is equal to the probability of finding the particle at

position x:(a) Always

(b) Only if is positive.

(c) Only if is normalized.

(d) Only if is real.

Answer (c)

2. [10 marks] Write down the Hamiltonian for Lithium Hydride, LiHAnswer:

H = Te+ Vnn+ Vne+ Vee

= h2

2me2

1

h2

2me2

2

h2

2me2

3

h2

2me2

4

+ 3e2

4oRLiH

3e2

4or1Li

3e2

4or2Li

3e2

4or3Li

3e2

4or4Li

e2

4or1H

e2

4or2H

e2

4or3H

e2

4or4H

+ e2

4or12+

e2

4or13+

e2

4or1 4+

e2

4or2 3+

e2

4or2 4+

e2

4or3 4

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3. (a) [15 marks] Decide whether or not each of the following wavefunc-tions obeys the Pauli Exclusion Principle for electrons:

i. (1, 2) =px(1)px(2)(1) (2)Answer N

ii. (1, 2) =px(1)py(2)(1) (2)Answer N

iii. (1, 2) =px(1)px(2)(1) (2)Answer N

iv. (1, 2) = {px(1)py(2) +py(1)px(2)}(1) (2)Answer N

v. (1, 2) = {px(1)py(2)py(1)px(2)}(1) (2)Answer Y

vi. (1, 2) =px(1)px(2){(1) (2) + (1)(2)}Answer N

vii. (1, 2) =px(1)px(2){(1) (2) (1)(2)}Answer Y

viii. (1, 2) = {px(1)py(2) +py(1)px(2)} {(1) (2) + (1)(2)}Answer N

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ix. (1, 2) = {px(1)py(2) +py(1)px(2)} {(1) (2) (1)(2)}

Answer Y

(b) [15 marks] Starting from the orbital-product wavefunction for Lithium,

(1, 2, 3) = 1s(1)(1)1s(2)(2)2s(3)(3)

construct a wavefunction which obeys the Pauli exclusion princi-ple.

Answer

(1, 2, 3) = Det (1s1s2s)

= (show working) . . .

= 1s(1) 1s(2) 2s(3)

+ 1s(2) 1s(3) 2s(1)

+ 1s(3) 1s(1) 2s(2)

1s(1) 1s(3) 2s(2)

1s(3) 1s(2) 2s(1)

1s(2) 1s(1) 2s(3)

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4. (a) [5 marks] Plot the antibonding molecular orbital forH+2 along the

internuclear axis without worrying about normalization.Answer

(b) [5 marks] Plot the probability density, again neglecting normal-ization.

Answer

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(c) [15 marks] The wavefunction for the ground state ofH+2 may beapproximated as a linear combination of atomic orbitals 1sA and1sB ,

= 1sA+ 1sB

i. Express the electron density for H+2 in terms of the atomic

orbitals, without worrying about normalization.Answer

(r) = (r)(r)

= (1sA(r) + 1sB(r))(1sA(r) + 1sB(r))

= 1sA(r)1sA(r) + 1s

B(r)1sB(r) +

1sA(r)1sB(r) + 1s

B(r)1sA(r)+

ii. Express the electron density as a sum of the atomic densitiesand an additional molecular density, again without worryingabout normalization.

Answer

(r) = 1sA(r)1sA(r) + 1s

B(r)1sB(r) +

1sA(r)1sB(r) + 1s

B(r)1sA(r) +

= (atom A) + (atom B) +1sA(r)1sB(r) + 1s

B(r)1sA(r) +

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(d) [25 marks] Consider the two-electron molecule H+3, and assume

that the three nuclei form an equilateral triangle. Starting fromthe Hartree-Fock wavefunction forH+3 ,

(1, 2) = (1)(2){(1)(2) (1)(2)}

Expand the molecular orbital as a L.C.A.O. of atomic orbitals1sA, 1sB and 1sC,

= 1sA+ 1sB+ 1sC

By expressing in terms of the atomic orbitals show that theHartree-Fock wavefunction gives unphysical behaviour at long bond-

lengths.Answer

(1, 2) = (1)(2){(1)(2) (1)(2)}

= (1sA(1) + 1sB(1) + 1sC(1))(1sA(2) + 1sB(2) + 1sC(2)) {(1)(2) (1)

= 1sA(1) 1sA(2){(1)(2) (1)(2)}

+1sB(1) 1sB(2){(1)(2) (1)(2)}

+1sC(1) 1sC(2){(1)(2) (1)(2)}

+1sA(1) 1sB(2){(1)(2) (1)(2)}+1sB(1) 1sA(2){(1)(2) (1)(2)}

+1sA(1) 1sC(2){(1)(2) (1)(2)}

+1sC(1) 1sA(2){(1)(2) (1)(2)}

+1sB(1) 1sC(2){(1)(2) (1)(2)}

+1sC(1) 1sB(2){(1)(2) (1)(2)}

There are nine terms in the expansion of the wavefunction. The firstthree terms, A(1)A(2), B(1)B(2) and C(1)C(2), have both electronson the same atom, ie they describe the ionic configurations HA H

+B H

+C

etc. At long bond-lengths H+

3 would be expected to dissociate toHAHBH

+C etc, so the first three ionic configurations are unphysical

at long bond-lengths. The remaining six terms, A(1)B(2) etc describethe configurationHAHBH

+C etc and hence are physically reasonable at

long bondlengths.

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5. The electrons in a cyclic polyene can be modelled as particles ona ring. It is easiest to use polar coordinates r and . The Hamil-tonian for a particle of mass m moving around a ring of radius r is

H= h2

2m r2d2

d 2. The mass of an electron is 9.11 1031 kg, and the

radius of the ring is 109 m.

(a) [75 marks] The following problems are for a single electron ona single ring:

i. Draw a diagram showing the ring, a single electron, r and .

ii. Using the Hamiltonian above, explicitly calculate the eigen-

values and eigenvectors for a single electron on a ring. Do notuse any formulae from the data sheet.

Answer

The radius r is fixed, so the eigenfunctions are functions of,(). The eigenfunctions ofd2/d2 are ei, where is anunkown constant,

h2

2m r2d2

d 2ei =

h22

2m r2ei

The eigenvalue of the eigenfunctione i is h22

2mr2.

(An equally good alternative set of eigenfunctions is cos and sin .)What values can take? The wavefunction must be single-valued, ie when we go once around the ring the wavefunction

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must return to its initial value,

(0) =(2 )

and hence

2 = 2 n wheren = 0,1,2, . . .

ie= 0,1,2,3 . . .

(note that= 1 gives a different eigenfunction from = 1).In conclusion, the eigenfunctions and eigenvalues ofH are,

n() = ein wheren= 0,1,2, . . .

En = h2 n2

2m r2

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iii. What are the allowed values for the quantum number(s)?Answer

n= 0,1,2, . . .

iv. What are the wavelengths of the eigenfunctions correspondingto the lowest three eigenvalues?

Answer

When answering this question it helps to draw a circle orradius r. Then cut the circle and straighten it out into a lineof length 2r. Then draw the wavefunction vs circle length.

E0 =

E1 = 2r

E2 = r

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(b) [20 marks] Consider a system containing two non-interacting

copies of the above system, ie two widely separated rings each ofradius 109 mand each possessing a single electron. What are theeigenvalues and degeneracies of the lowest three eigenvalues?

Answer

Because the two systems are non-interacting, the Hamiltonian isseparable. So the energy of the combined system is the sum of theenergies of the individual rings.

The ground state energy is E0 0 = 0. The ground-state eigen-function is four-fold degenerate; the four degenerate states are0A 0B, 0A 0B, 0A 0Band 0A 0B, where A and B de-notes ring A and ring B.

The first excited state energy is E0 1 = E1 0 = E01 = E1 0 =h2

2mr2. . The first-excited-state eigenfunction is sixteen-fold de-

generate; the sixteen degenerate states are 0A 1B, 0A 1B,0A1B, 0A1B, and so on for 10, 0 1 and1 0.

The second excited state energy is E1 1 = E11 = E11 =E1 1= 2

h2

2mr2. This state is also sixteen-fold degenerate.

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(c) [30 marks] Consider four electrons on a single ring. For

simplicity, treat the four electrons as independent particles, ieignore Coulomb interactions between the electrons.

i. Redraw the energy-level diagram for the lowest five energylevels of a single electron, and indicate which levels are occu-pied in the ground state by using up-arrows for electrons with spin and down-arrows for electrons with spin.

ii. What is the total energy of the ground state?Answer

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E0(1)0(2)1(3)1(4) = 2 h2

2mr2.

iii. Is the ground state a singlet or a triplet?Answer

Triplet (Hunds rules: not on the syllabus in Spring 2002).

iv. Calculate the absorption frequency for a transition from theground state to the first excited state.

Answer

The lowest-energy transition is marked on the diagram,

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E= E1 E0 = h2

2m r2 0

To convert this to a wavelength use

E= h=hc/

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(d) [30 marks] Consider a single spinless particle on a single

ring. Using the normalizedwavefunction,

() = 1

18( 2)

calculate the average value for the energy. Is this above or belowthe true ground-state energy?

Answer

< H > =

H

=

H since is normalised

= 20

H

= h2

2 m r2

20

d2

d 2

= h2

2 m r2

20

1

18

= h2

2 m r2

20

1

18(2 2)

1

18

= h2

2 m r21

182[3/3 2]20

= h2

2 m r21

182

8 3/3 43

= h2

2 m r21

1824 3

3

= h2

2 m r21

1824 3

3

According to the variational theorem, this energy is above the true

ground-state energy. This is easy to verify in the present case sincethe true ground-state energy is zero.

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(e) [40 marks] Consider a single spinless particle on a single

ring in the first excited state.i. What is the degeneracy of this state? For each of the de-

generate states, calculate the average value of the angular-momentum operator,

J=h

i

d

d

Answer

The degeneracy of this state is two; the two states are e i andei . The average value ofJfor the first state is,

< J > =

2 0 e

i hidd

ei20 e

i ei

=

2 0 e

i hidd

ei20 e

iei

=

2 0 e

i hi

iei20 e

iei

=

2 0 h

20 1

= h

So the average value ofJfor the first state is h. Proceedingsimilar for the second state, one finds that the average valueofJ ish.

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ii. Consider a complex linear combination of the degenerate

states, eg, = state1+ i state2

Evaluate the average value of the angular momentum operatorfor , and also for .

hint: To shorten the computation, use the fact that state 1and state2 are orthogonal.

< J > =

2 0 (1+ i2)

J(1+ i2)2 0 (1+ i2)

(1+ i2)

=2 0

1 J 1+ (i2) J i2+ 1 J i2+ (i2)

J 120

11+ (i2) i2+ 1i2+ (i2)

1

=

2 0 h + (i)i(h) + i

1 J 2 i

2 J 120

11+ (i2) i2+ 0 + 0

=

2 0 h h+ ie

iJ ei ieiJ ei

1 + 1 + +0 + 0

=

2 0 +ie

i hidd

ei iei hidd

ei

4

=

2 0 +ie

i hi(i)ei iei h

iiei

4

=2 0 iei2h iei2h

4

=ih[ e

i2

i2 + e

i2

i2 ]20

4 buteix has period 2 so each expontential in the numerator gives ze

= 0

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Spare page

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2

1. (6 pts) In an X-ray photoelectron experiment, a photon of

wavelength 150 pm ejects an electron from the inner shell of an atom

and it emerges with a speed of 2.14x107m/s. Calculate the binding

energy (= work function) for this electron (in eV).

2.(6 pts) Is the function (x) = cos(kx) + sin(kx) an eigenfunction

of the operator = d2/dx2? If the answer is "YES", what is the

eigenvalue? (Show work)

OLD EXAM Solutions

i) Calculate the kinetic energy of the electron ( = 1/2 m v^2 )ii) Calculate the energy of the photon ( E = h nu)iii)Binding energy = Photon Energy - Kinetic Energy

Ionization Threshhold

Bound Electron

Laser ExcitationKinetic Energy

i) d^2/d x^2 psi = - k^2 cos(kx) - k^2 sin(kx)

The wavefunction is an eigenfunction of the second derivative operator, with eigenvalue -k^2


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3

3. (6 pts) Given () = Nexp(-im) with 0 2, m = integer, find

the factor N that normalizes ().

4.(10 pts) The energy (E) of the classical one-dimensional

harmonic oscillator is given by the expression

E = px2/2m + 0.5kx2

where k = force constant; m = mass; and px= momentum.

(a) (5 pts) What is "x" in the above expression? [A one sentence

definition will do.]

(b)(5 pts) Write the expression for the quantum mechanical

harmonic oscillator Hamiltonian

N = 1/sqrt(2 pi)

x is the position of the particle; in classical mechanics x(t) is thetrajectory.

H = -hbar^2/2m d^2/dx^2 + 0.5 k x^2


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4

5.(6 pts) The energy separation between the vibrational levels in a14N2molecule is found to be 4.8x10

-20J. What is the zero-point

vibrational energy for a 14N2molecule (in J)?

6.(12 pts total) Let 1 and 2be two normalized eigenfunctions of ahermitian operator with eigenvalues 1and 2, respectively. Assume

1and

2are different (

1

2)

a. (6 pts) Evaluate the integral 2* 2 d

Give each step of the evaluation clearly. You will not receive any

credit, if you only show (guess) the answer.

b. (6 pts) Evaluate the integral 1* 2 d

Give each step of the evaluation clearly. You will not receive any

credit, if you only show (guess) the answer.

E_n = h nu( n + 1/2)

spacing = E_n - E_n-1 = h nu

zero point = 1/2 h nu

So zero point energy = 1/2 spacing

int psi^* O psi = int psi^* E psi = E int psi^* psi = E

So the inegral equals the eigenvalue.

int psi_1^* O psi_2 = int psi_1^* E_2 psi_2 = E_2 int psi_1^* psi_2

= 0 since the eigenfunctions of a hermitean operator are orthogona


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5

7. (6 pts) The magnitude of the total angular momentum is measured for

a particle with the result |L| = 6h

If Lzis now measured for this particle, what is (are) the

possible value(s) obtainable?

8.(30 pts total) A particle of mass m in a one-dimensional box of

length L is in the state characterized by n = 3.

a. (8 pts) Draw a sketch of the wavefunction and a sketch of the

the probability density for this particle (label the axes!).

b. (4 pts) What is(are) the most likely location(s) of this particle?

|L|^2 = 6 hbar^2 = l(l+1) hbar^2

So the angular momentum quantum number l = 2

The z component is m hbar, where m is between -2 and 2.

So the values obtainable are -2hbar, -hbar, 0, hbar, 2hbar.

WavefunctionDensity

The most likely locations occur where the density is at a maximum,ie at L/6, L/2, 5L/6

1/3L 2/3L

0 L


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6

c. (6 pts) What is the probability of finding this particle in the

central third of the box?

d. (6 pts) For this particle, what is ?

e. (6 pts) For this particle, what is ?

The integral of the density from 0 to L must equal one.Looking at the density, it is clear by symmetry that the integral over the middle third of the box isone third of the total, ie one third of one, ie 1/3.

Answer: Probability = 1/3.

= E_3 = hbar^2 n^2 3^2/2 m L^2

= 0

The easiest way to see this is

= int psi^* p_x psi= int even odd even= int odd since odd times even = odd= 0 by symmetry

Alternatively, you can actually evaluate the integral.


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7

9.(12 pts) A particle of mass m and moment of inertia I rotates freely

in three dimensions in a state described by the spherical harmonic

Y3,-2(,)

(a) (6 pts) Find the total rotational energy of this particle

(b) (6 pts) In a single measurement of the z-component of the angular

momentum for this particle, what is the probability of obtaining the

value 1h?

10.(6 pts) What are the values of s and msfor an electron in the

spin state ?

H = L^2/2 I

Y is an eigenfunction of L^2 with eigenvalue hbar^2 3(3+1)

So Y is also an eigenfunction of H, with eigenvalue hbar^2 3(3+1)/2 I

So the energy is hbar^2 3(3+1)/2 I

The possible values are 3hbar, 2hbar, 1hbar, 0, -1hbar, -2hbar, -3hbar.

All 7 energy levels are degenerate and hence equally probable.

So the probability of getting 1hbar is 1/7.

s = 1/2, m_s = -1/2


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8

Operators

H = (-h2/2m)2+ V

2 = 2/x2+ 2/y2+ 2/z2 = 2/r2+ (2/r)/r + (1/r2)2 =

2/r2+ (2/r)/r - (1/r2)(L2/h2)

2 = (1/sin2)2/ 2 + (1/sin)/sin/

H = (-h2/2I)2/ 2

H = (-h2/2m)d2/dx2

Energies Wavefunctions

En= n2h2/8mL2 n = 1,2,3,4 n(x) = (2/L)1/2sin(nx/L)

El = l (l +1)(h2/2I) l = 0,1,2,3.. Yl ,ml (,) = l ,ml ()ml ()

Ev= (v + 0.5)h v = 0,1,2,3.. v(x) = NvHv(y)exp(-y2/2)

E = J2/2I

E = h

E = 0.5mv2= p2/2m

Miscellaneous

H= E = d ijd = ij

L2Yl ,ml (,) = l (l +1)h2Yl ,ml (,) l = 0,1,2,3,4..

LzYl ,ml (,) = mlhYl ,ml (,) ml = 0,1,2..l

pq h/2 = c = h/p

[1,2] = 12- 21 = mAmB/(mA+ mB)

Fundamental constants and conversion factors

h = 6.626x10-34Js ao = 0.529x10-10 m c = 3.00x108m/s

1 eV = 1.60x10-19J me= 9.11x10-31kg 1 J = 1 kgm2/s2

1 pm = 1.0x10-12m

Integrals

sin(nx/L)dx = (-L/n)cos(nx/L)

sin2(nx/L)dx = 0.5x - (L/4n)sin(2nx/L)

sin(nx/L)cos(nx/L)dx = (L/2n)sin2(nx/L)


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Past Papers Solutions Output