Giving or receiving aid is cause for dismissal from the University. You may have one two-sided, 8-1/2x11-inch page of notes. Calculators may be used, but not computers. Show all work for Short and Long Problems on the exam booklet. No other materials may be used during the exam. The exam is 2 hours long.

E = gm c² = Ö[( m² c⁴ + c² p²)] K = E - mc² p = gm v c p/ E = v/c

E_photon = cp_photon               c = lf

p = h/l = (^h/_2p) k E = hf = (^h/_2p)w Dx Dp ³ (^h/_2p)/2 DE Dt ³ (^h/_2p)/2

-[((^h/_2p)²)/2 m] [(d²)/(dx²)] y(x) + U(x)y(x) = E y(x)

E_n = [(n² h²)/(8 m L²)] = [(n² p² (^h/_2p)²)/(2 m L²)]              y_n(x) µ sin(( n px)/L )

E_n = -Z² hcR/n²

K_out - K_in = (M_in - M_out)c²

N(t) = N(0) e^{-r t}             t = 1/r             t_1/2 = ln2 / r

c = 3.00 ×10⁸ m/sec e = 1.60×10^-19 C h = 6.63×10^-34 J·sec
1 eV = 1.60×10^-19 J (^h/_2p) = h/2p = 1.06×10^-34 J·sec
hc = 1240 eV·nm (^h/_2p) c = 197 eV·nm R = 0.0110 nm^-1 hcR = 13.6 eV

m_e = 0.511 MeV/c² = 9.11 ×10^-31 kg
m_p = 938.3 MeV/c² = 1.673 ×10^-27 kg
m_n = 939.6 MeV/c² = 1.675 ×10^-27 kg
m_n = 0 (neutrino)
1 u = 931.6 MeV/c²

I. Ten Multiple Choice Problems (30 points total, each 3 points, no part marks)

Indicate the correct answer.

1. What is the typical binding energy of a valence electron to a neutral atom? (i.e. to within a factor of 5)
(a) 10 eV
(b) 10 keV
(c) 10 MeV
(d) 10 GeV
(e) 10 TeV

2. What is the typical binding energy per nucleon for a stable nucleus?
(a) 10 eV
(b) 10 keV
(c) 10 MeV
(d) 10 GeV
(e) 10 TeV

3. Which of the following atomic electronic configurations is prohibited by the Pauli exclusion principle?
(a) 1s² 2s² 2p⁶ 3s²
(b) 1s² 2s² 2p⁶ 4s²
(c) 1s² 3s² 3p¹ 4s²
(d) 1s¹ 2s³ 2p⁶ 3s¹
(e) none of the above

4. Consider the following stable isotopes of lithium, iron and lead, ⁷₃Li, ⁵⁶₂₆Fe and ²⁰⁸₈₂Pb. Which is the correct ordering of the nuclear binding energy per nucleon (B/A)?

(a) Li < Fe < Pb
(b) Pb < Fe < Li
(c) Fe < Pb < Li
(d) Li < Pb < Fe
(e) none of the above

4. Suppose a given nucleus has a lifetime of t = 722 sec. What fraction of a sample of these nuclei will still be around, undecayed, after 1444 sec? (there are two 4's so this is a bonus question)

(a) 0.07
(b) 0.14
(c) 0.50
(d) 0.73
(e) none of the above

5. What is the minimum wavelength photon that can be emitted from an excited Li²⁺ ion (to ±1 nm accuracy)
(a) 10 nm
(b) 23 nm
(c) 55 nm
(d) 91 nm
(e) none of the above

6. Which of the following transitions is allowed for a hydrogen atom?

(a) 3s ® 1s
(b) 3d ® 1s
(c) 3p ® 2p
(d) 3s ® 2p
(e) none of the above

7. What is the spin of a neutron (s = ?)?

(a) 0
(b) 1/2
(c) 1
(d) 2
(e) none of the above

8. Which of the following nuclear reactions is prohibited by a conservation law?

(a) p + ¹³C ® n + ¹³N
(b) n ® p + e^- + [`(n)]
(c) ²³⁶U ® ²³⁵U + n
(d) ²H + ²H ® ³He + n
(e) none of the above (all are permitted)

9. No nuclei occur naturally on the earth which have atomic numbers larger than Z = 92. Which force is mainly responsible for destabilizing nuclei with larger atomic numbers?
(a) gravity
(b) electromagnetism
(c) nuclear (strong) force
(d) weak force
(e) none of the above

10. Consider the one electron in a neutral hydrogen atom. If the electron has total angular momentum quantum number l = 2, which of the following is certainly not true?
(a) the orbital angular momentum of the electron is Ö6(^h/_2p)
(b) the z-component of orbital angular momentum is L_z = 0
(c) the electron is in its ground state
(d) another electron could be added to the same orbital
(e) the electron has z-component of spin S_z = (^h/_2p)/2

II. Two Short Problems (30 points total, each 15 points)

11. Consider a sodium atom (²³Na).

(a) Write down the ground-state electronic configuration using the standard notation (e.g. 1s²¼) (2 points)

(b) Write down the electronic configuration of the first excited state. (2 points)

(c) Using hydrogen-like energy levels, estimate the energy difference between the second excited state and the third excited state (4 points)

(d) Suppose we knock one of the 1s electrons out of the atom, allowing a 2p ® 1s transition to occur into the vacant 1s state. Use hydrogen-like energy levels to estimate the energy of the photon that is emitted (4 points)

(e) Is a crystal of Na atoms a metal (i.e. electrically conductive)? Explain your answer in terms of energy bands (3 points)

12. Consider the quantum mechanics of a particle of mass m that can move along the x-axis, with potential energy U = kx²/2. The ground state has the form y₀(x) = exp[ -ax²/2].

(a) Write the Schrodinger equation, determine what a has to be in order for y₀ to solve it, and therefore determine the ground state energy E₀ in terms of (^h/_2p), m and k (6 points)

(b) Sketch the ground state wavefunction (easy, since its form is given in (a)) and the first two excited states as a function of x, on the same graph below (6 points)

(c) Explain in one or two sentences how the Schrodinger equation of (a) could be used to describe small deformations of a diatomic molecule like H₂ (3 points).

III. Two Longer Problems (40 points total, each 20 points)

13. (a) Calculate the nuclear binding energy per nucleon for the nuclei ¹²C, ⁵⁶Fe, and ²³⁸U, exactly, from the atomic mass data given (9 marks)

(b) Draw a graph of binding energy per nucleon B/A versus nucleon number A, indicating your results of part (a) (3 marks)

(c) What is the approximate range of nucleon number A for which exothermic fusion of the symmetrical form ^AX + ^AX ® ^2AY can occur ? (2 marks)

(d) What is the approximate range of nucleon number for which exothermic fission of the symmetrical form ^AX ® ^A/2Y + ^A/2Y can occur? (2 marks)

(e) Write the reaction for the b^--decay of ⁵⁶Mn, identifying all the decay products, and determine the kinetic energy carried off by the decay products (4 marks)

14. Our solar system is made up of nuclei that were created during the explosion (supernova) of another star a long time ago. In that explosion, equal numbers of ²³⁵U and ²³⁸U nuclei (N_0,235 = N_0,238 = N₀) were created. These uranium isotopes have been decaying ever since. It turns out that ²³⁵U decays faster, so it is the rarer isotope.

(a) Write down the reactions for a decay of ²³⁵U and ²³⁸U, clearly identifying the product nuclei (4 points)

(b) Determine the kinetic energy released in the a decays of (a) (6 points)

(c) Write down formulae for N₂₃₅(t) and N₂₃₈(t) a time t after the explosion, in terms of the decay rates r₂₃₅ and r₂₃₈ (2 points)

(d) Use the results of (c) to write a formula for the ratio of ²³⁵U to ²³⁸U as a function of time, i.e. N₂₃₅(t)/N₂₃₈(t) (2 points)

(e) On earth now, the ratio of ²³⁵U to ²³⁸U is N₂₃₅/N₂₃₈ = 0.0072. The half-lives are 704×10⁶ years for ²³⁵U, and 4.47×10⁹ years for ²³⁸U. How many years ago did the supernova that created our uranium explode? (6 points)

Solutions

2. (c) recall that B/A is around 8 to 9 MeV for most nuclei

3. (d) s-orbitals can hold at most 2 electrons

4. (d) iron is the most bound, heavy elements such as Pb have B/A around 8 MeV, light elements like Li have much lower B/A values due to their large amount of surface area relative to their volume

4. (b) N(1444     sec) = N(0) exp[-1444    sec/722     sec] = 0.14 N(0)

(note that this was a bonus question, making the total out of 103 marks!)

5. (a) the maximum energy photon (with minimum wavelength) that can be emitted from a one-electron ion comes from a transition from a very large-n state, to n = 1, and is DE = -Z² hcR. For Li, Z = 3, DE = 122.4 eV. The wavelength is l = hc/DE = 1240 / 122.4 = 10.1 nm.

6. (d) we must have Dl = 1, so s-s, d-s, and p-p transitions are out, the s-p transition is the only one allowed.

7. (b) protons and neutrons are fermions with spin quantum number s = 1/2.

8. (e) all the reactions conserve electric charge and nucleon number, so all are allowed.

9. (b) Coulomb self-repulsion of the large number of protons destabilizes large nuclei

10. (c) in the ground state l = 0 is the only possibility.

(a) 1s² 2s² 2p⁶ 3s¹

(b) The first excited state is of the outermost 3s¹ valence electron into the nearest higher energy level, or into 3p¹, to give the electronic structure
1s² 2s² 2p⁶ 3p¹

(c) The second excited state is the 3d¹ state (note that these are hydrogen-like excited states, and the filling order of many-electron atomic orbitals is irrelevant), and the third excited state is the 4s¹ state. Since the outer electron `sees' an effective nuclear charge of +e (there are 10 electrons between it and the nucleus of charge +11e) these two states are spaced by just the hydrogenic spacing

(d) The inner electrons see essentially the entire nuclear charge of +Ze = +11e, so the 2p® 1s transition releases about

(e) A crystal of Na is an electrically conductive metal since its valence band (3s) is half-filled.

-(^h/_2p)² a²/(2m) + k/2 = 0 or a² = mk/(^h/_2p)²
The remaining term determines the energy:
E = (^h/_2p)² a/ (2m), into which we plug a = (mk/(^h/_2p)²)^1/2 to obtain

(b) See the figure at the end of this document.

(c) When the length of the bond in H₂ is at its equilibrium length, it takes its lowest energy. Therefore for small deformations x of the bond length away from its equilibrium length, the potential energy behaves as k x²/2 for some k. So the Schrodinger equation of (a) could be used to treat molecular vibrational motion.

For ¹²C we have Z = 6, N = 6 and A = 12, so
B = (6×1.0078 + 6×1.0087 - 12.0000) uc² ×931.5 MeV/(u·c²) = 92.2 MeV
or B/A = 7.7 MeV

For ⁵⁶Fe we have Z = 26, N = 30 and A = 56, so
B = (26×1.0078 + 30×1.0087 - 55.9349)×931.5 MeV = 492.7 MeV
or B/A = 8.8 MeV

For ²³⁸U we have Z = 92, N = 146 and A = 238, so
B = (92×1.0078 + 146×1.0087 - 238.0508) ×931.5 MeV = 1804.3 MeV
or B/A = 7.6 MeV

(b) See attached graph. You should show clearly that ⁵⁶Fe is at the peak of B/A.

(c) We need 2A < 56 so we must have A < 28, roughly.

(d) We need A/2 > 56 so we must have A > 112, roughly.

(e) Since Z = 25 for Mn, emission of an electron will push the charge of the final nucleus to Z = 26, so the reaction is ⁵⁶Mn®⁵⁶Fe+ e^- + [`(n)]

The KE released is easily computed from the mass difference of inputs and outputs:
DK = ( m_{⁵⁶Mn,nucleus} - m_{⁵⁶Fe,nucleus}- m_e - m_n )c²

If we add 25 electron masses to the inputs and outputs, and note that m_n = 0 we can write

DK = ( m_{⁵⁶Mn,atom} - m_{⁵⁶Fe,atom} )c²

and plugging in the atom mass data from Appendix D we obtain
DK = ( 55.9389 - 55.9349 ) ×931.5 MeV
giving DK = 3.72 MeV

(b) DK = (m_{U-235,nucleus} - m_{Th-231,nucleus} - m_He-4,nucleus ) c²
Again we write in terms of atom masses by adding and subtracting 235 or 238 electrons:
= (m_U,atom - m_Th,atom - m_He,atom )

For ²³⁵U:
DK = (235.0439 - 231.0363 - 4.0026)×931.5 = 4.7 MeV

For ²³⁸U:
DK = (238.0508 - 234.043 - 4.0026)×931.5 = 4.3 MeV

(c) The decays are each exponential, and start at N₀ nuclei at time t = 0, but with different decay rates:
N₂₃₅(t) = N₀ e^{-r₂₃₅ t}
N₂₃₈(t) = N₀ e^{-r₂₃₈ t}

(d) Just divide the results of (c):

(e) Now we need the numerical rates:
r₂₃₅ = ln2 / 704×10⁶ yr = 9.8×10^-10 yr^-1

r₂₃₈ = ln2 / 4.47×10⁹ yr = 1.6×10^-10 yr^-1

So, r₂₃₅ - r₂₃₈ = 8.2 ×10^-10 yr^-1

Now we rewrite the result of (d) as