REAL TEST 4; NOTE: HERE'S A LINK TO FINAL EXAM PART 2. After you read these solutions, given at the end of each problem under heading 'SOLUTIONS," you might want to take advantage of it. It's your call. There were general student errors on #6 , #8 and 9(g) for example. HOWEVER, THE OVERALL PERFORMANCE WAS EXCELLENT CONSIDERING THE HEAVY OVERLOADS AND EXCESSIVE PRESSURE BURDENING YOUR EMOTIONS AND COGNITION. So I am generally satisfied with scores and make no apologies the final was a complete take home, which with your feedback averted an unmitigated academic catastrophe under my watch. Let it be known my teaching philosophy is simple: All things considered, I try my best to test students on what they know, not on what they don't know; so now you know. Good luck in your future academic and employment endeavors!

1. CH. 39. WAVE FUNCTION REQUIREMENTS
A particle is described by the wave function

0 for x < 0

Ce ^{- x/L}for x 0, where L = 1 nm.
(a) (10 points) Find the constant C using that fact that the particle must be found somewhere on the x-axis with probability 1. In other words , normalize the wave function!
(b) (10 points) Draw graphs of and the probability density .
(c) ( 10 points) Calculate the probability of finding the particle in a region x 1 nm.
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SOLUTIONS:

(a) INTEGRAL OF C²e^-2x/Ldx from 0 to infinity = 1 or C²*(L/2)[1 - 0] = 1

Thus C = sqrt(2/L) = 4.47x10 ¹² m ^-1, a classic result.
(b) Graphs are exponential with a more drastic rate of change due to doubling the exponent to -2x/L.
(c) INTEGRAL OF (2/L^)*e-^2x/Ldx from 1 nm to infinity
= (2/L)*(L/2)*[e-^2x/L - 1]= [e-^2x/L - 0] = 0.135, where x = 1 nm.

2. CH. 40. TUNNELING
(a) (20 POINTS) An electron with initial kinetic energy 32 eV encounters a "square" barrier with height 41 eV. The probability the electron tunnels through the barrier is 0.13 %. What is the barrier width in nm? Be mindful of converting the percentage to a decimal before beginning calculations.
(b) ( 5 POINTS) Suppose a proton with the same kinetic energy encounters the same barrier. What is the probability the proton will tunnel through the barrier? You may express your answer in terms of some base raised to a negative power without evaluating the expression. Does your answer seem reasonable?
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SOLUTIONS:

(a) 0.0013 = Ge^-2k'L , E/Uo = 32/41 = 0.780 in the formula for G. Thus: L = 0.25 nm.
(b) T = Ge^-2k'L = 10 ^-143 , an expected result since the wavelength of the proton is so much smaller, making scattering effects very small. NOTE: Ge-^2k'L = 10 ^-X , WHERE X = -Log₁₀ (Ge-^2k'L ), for L = 0.25 nm and a much larger wave number k' , which is proportional to the sqrt (mass); the proton has a much larger mass than the electron's, making k' much much larger than k..

3. CH. 40. STEP FUNCTION POTENTIAL DESCRIBED BY #35, QUIZ 7.

Particles incident from the left are confronted with a step in potential energy. See #35, Quiz 7. Classically, all the particles would pass into the region of higher potential. But according to Quantum Mechanics, a fraction of the particles are reflected at the barrier. This problem is analogous to the partial reflection and transmission of light striking an interface between two different media covered in Physics 4C.

Suppose a particle with kinetic energy E = 7.00 eV is incident from a region where the potential energy is zero onto one in which U_o = 3.50 eV.

(a) (15 POINTS) Find the fraction of the incident particles that are reflected.
(b) (5 POINTS) Find the fraction of the incident particles that are transmitted "over" barrier.

HINT:

(a) USE THE RESULT OF #35, Quiz 7, AND COMPUTE |B|²/|A|²
(b) This quantity is easy to find since:

fraction of reflected particles + fraction of transmitted particles = 1. You can also compute |C|²/ |A|² directly; the two methods should give the same result if you work carefully---carefully in the sense that if you use |C|²/ |A|² directly, you have to to do the following: Multiply |C|²/ |A|² by (v'/v), the ratio of the "speeds" of the particles in the region x > 0 and x <0, respectively. That's because the probabilities are ratios of "fluxes" of particle given by speed*

probability current, sometimes called probability "flux."

NOTE, IN THE CASE OF A PLANE WAVE: speed = h_bar *k/m in general, where k is the wave number = 2*pi/lambda. In the case of the probability of reflection, the "speed ratio" was 1 since the speed of the reflected particles is the same as the speed of incident particles; that's because the particles are traveling in opposite directions in the SAME region. FOR MORE INFO GO HERE: http://en.wikipedia.org/wiki/Probability_current and scroll down to the section on

'Transmission and reflection through potentials' and also

'Examples' dealing with 'Plane wave' just below it.

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SOLUTIONS:
(A) A+ B = C and continuous derivative leads to
(B) ik*(A - B) = ik'*C, or
k*(A - B) = k'*C, where k and k' are the wave numbers x < 0 and x > 0, respectively.
Solve for B and C in terms of A. Reproduce the answer in back of book.

A = (1/2)*[1   + k'/k]*C = (1/2)[k + k'](C/k); thus C = 2Ak/[k + k']

and

B = (1/2)*[1 - k'/k]*C = (1/2)*[1 - k'/k]*2Ak/[k + k'] =
(1/2)*[k - k']*2A/[k + k'] =A[k - k']/[k + k '] ; thus, B = A[k - k']/[k + k '] .

Thus:    C/A = 2k/[k + k'] (NOTE: DO NOT USE DIRECTLY UNTIL YOU READ THE ABOVE NOTE ON "FLUXES")

and B/A = [k - k']/[k + k '].

NOTE: k = sqrt(2)*k' since E = 2*Uo .

B/A = [sqrt(2)*k' - k']/[sqrt(2)*k' + k '] = (0.41)/[1.41 + 1] = 0.1701.

THUS: fraction reflected = |B|²/|A|² = 0.0292 = or 3 % and thus fraction transmitted = 1 - 0.0292 = 0.971 or 97 %

4. CHAPTER 41: THE HYDROGEN ATOM: IT MAY BE HELPFUL TO USE THE WAVELENGTH EXACTLY AS WRITTEN , KEEPING ALL THE SIGNIFICANT DIGITS GIVEN.

(23 points) A hydrogen atom in the 2nd excited state (n =3) absorbs an infra-red photon of wavelength 1874.606 nm.

(a) (10 points) Determine the maximum possible orbital angular momentum of the electron after the absorption. Note: The orbital angular momentum is given by the formula:
.
(b) (10 points) Assume the orbital angular momentum of the final state (after absorption) has the maximum value you computed in part (a). What must be the orbital angular momentum of the initial state (n = 3)? Is the initial state an s-state, p-state, d-state or f state? Circle one.
(c) (3 points) Suppose the final state has L_z with quantum number m_l = 1. What are the possible values of m_l for the initial state?
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SOLUTIONS:
(a) hc/lambda = R*[1/3² - 1/n_f²], where

$R_\infty = \frac{m_{\mathrm{e}} e^4}{8 \varepsilon_0^2 h^3 c} = 1.0973\;731\;568\;539(55) \times 10^7 \ \mathrm{m}^{-1},$

Solving, n_f = 4. That means l_MAX = 3, and thus L_MAX = sqrt(12)*h_bar
(b) SELECTION RULES are such that the
(change in L) = +1 or -1
and
(change in m_l ) = 0, -1 or +1.
Thus: n=3 state has L = 2, a d state.
(c) ASSUME: For the final state (n = 4, L= 3), m_l = 1. Since (change in m_l ) = 0, -1 or +1, the state (L = 2) has initial m_l = 2, 1, 0. NOTE: FOR THE INITIAL STATE m_l = cannot be -1 or -2.

5. (25 points) HYDROGEN ATOM:
Consider the ground-state wave function for hydrogen below and in the textbook.
(a) (10 points) Show the wave function as written is normalized.
(b) (10 points) Find the probability of locating the electron between r₁ = a_o/2 and r₂ = 3a_o/2.

(c) (5 points)

Based on your answer to part (b) what is the probability the electron is found outside the region of part (b)?

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SOLUTIONS:

(a) Go here: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydrng.html#c1
You will see the function is normalized by making the lower limit zero and the upper limit infinity. SEE ALSO EXAMPLE 41.3

(b) SEE EXAMPLE 41.3 for guidance. Integration needed but you may use the indefinite integral formula or perform yourself using integration by parts. Evaluate the indefinite integral formula AT LIMITS r₁ = a_o/2 and r₂ = 3a_o/2 and compute the difference to get definite integral = 49.7 %

(c) 1.000 - 0.497 = 0.503 = 50.3 %

6. CH. 41 ZEEMAN EFFECT:

A hydrogen atom undergoes a transition from a 3d state to a 2p state.

(a) ( 5 points) In the absence of a magnetic field, what is the wavelength of the photon emitted? Use as many significant figures as possible from published data.

The atom is then placed in an magnetic field in the z-direction. Ignore spin effects. Consider only the interaction between the magnetic field and the atom's orbital angular momentum.

(b) (12 points) How many different wavelengths are observed for the 3d to 2p transition? Make a concise list of the m_l values for the initial and final states for the transition that leads to each photon wavelength. The table below may have more rows then the number of transitions. I am not asking you to compute wavelengths; just list the transitions according to initial and final m_l.

transition	initial m_l (3d)	final m_l (2p)
shorter wavelength	2	1
shorter wavelength	1	0
no change in wavelength	1	1
longer wavelength	0	1
no change in wavelength	0	0
shorter wavelength	0	-1
no change in wavelength	-1	-1
longer wavelength	-1	0
longer wavelength	-2	-1

(c) (6 points) Two observed wavelengths are exactly the same with the magnetic field as without. What are the initial and final m_l values for each transition that produces a photon of this wavelength?
(d) ( 4 points) Two observed wavelengths with this field are longer than the wavelength without the field.   What are the initial and final m_l values for each transition that produces a photon of this wavelength?
(e) (4 points) Two observed wavelengths with this field are shorter than the wavelength without the field.   What are the initial and final m_l values for each transition that produces a photon of this wavelength?
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SOLUTIONS:
(a)
hc/lambda = R*[1/3² - 1/2²], so lambda = 656.3 nm.

(b) SEE TABLE ABOVE FILLED ACCORDING TO SELECTION RULES:
(change in L) = +1 or -1; THERE ARE 9 POSSIBLE TRANSITIONS. THERE ARE MORE ROWS THAN TRANSITIONS AS MENTIONED.
and (change in m_l ) = 0, -1 or +1.

(c) ACTUALLY, THERE ARE 3 transitions if you include m_l = 0 to m_l = 0. THE OTHER TWO TRANSITIONS ALSO HAVE THE SAME m_l in the initial and final state; They are all shown in table above with the label " no change in wavelength." You were asked to pick two of the three.

(d) ACTUALLY, THERE ARE 3 such transitions; see table above. You were asked to pick two of the three.

(e) ACTUALLY, THERE ARE 3 such transitions; see table above. You were asked to pick two of the three.

7. (15 points) XRAY PHOTONS:

A K-alpha X-RAY is emitted from a sample with energy 1.2536 KeV .

(a) (10 points) Of which element is the sample made?
(b) (5 points) List the steps leading to the emission of the x-ray photon. Include why the photon was emitted and the atomic transitions leading to the photon emission. What was the initial state of the transition? What was the final state? Why was there a vacancy in the final state allowing it to be filled by an electron from a higher energy state? Also explain key factors in the formula you used to find information identifying the element; why is (Z -1) used in the formula ?
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SOLUTIONS:
(a) See equation   41.30: 1.2536x10³ eV = (Z -1)²*(10.2 eV); this yields Z = 12, for Mg aka magnesium.
(b) Electron from a beam knocks an atomic electron out of the n =1 orbital and another electron "falls" from an atomic n = 2 state into the vacancy. The "falling" electron sees a nucleus partially shield by the other electron in the n = 1 state .

8. MOLECULAR SPECTRA

(a) (10 points) For NaCl, three successive wavelengths for rotational transitions are 23.1 mm, 11.6 mm, and 7.71 mm. Calculate the moment of inertia of NaCl.
(b) (5 points) What is the bond length of NaCl?
(c) (10 points) Photons of what frequency can be spontaneously emitted by CO molecules in the state of vibrational number n = 1 and rotational number    l = 0?
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SOLUTIONS:
(a) THIS WAS THE MOST DIFFICULT PROBLEM OF THE ENTIRE ASSIGNMENT, JUDGING FROM STUDENT ANSWERS SURVEYED; that is why your extra credit opportunities in Ch. 43 may help a little.
Erot = l(l +1)*h_bar² /(2I). The ALLOWED CHANGE in energy is between   1 and l + 1 or l and l - 1, SINCE A PHOTON CARRIES AWAY (OR GIVES) ONE UNIT OF ANGULAR MOMENTUM.

In either case, the difference may be algebraically shown to be
E(l +1) - E(l) =
(h_bar²*l)/I, where l = 1, 2, 3, 4, ...... NOTE: l = 1 in the series refers to a transition from   l = 1 to l = 0. We see the slope of the line is h_bar ²/I and the change in energy is linearly proportional to l:

slope = h_bar ²/I = [hc/lambda₂   - hc/lambda₁ ]/(change in l),    where you may choose two lambdas, designated as lambda₁    and    lambda₂ . Remember to divide the numerator by the (change in l).
FOR EXAMPLE, LET lambda₁ = 23.1x10 ^-3 m and lambda₂ = 11.6x10 ^-3 m ; in this case we see the (change in l) = 1, since the wavelengths are adjacent and represent successive transitions which change l by one, the selection rule for rotational state changes:

hc/(11.6x10 ^-3 m) - hc/(23.1x10 ^-3 m)   = 0.854x10^-23 J =   h_bar ²/I . This gives I = 1.29x 10^-45 kg-m².
(b) I is given by formula 42.5: I = m_rr_o² , where = m_r = m1*m2/(m1 + m2). This gives   = 2.36x10^-10 m.

(c) WE TURN OUR ATTENTION TO CARBON MONOXIDE, THAT NOXIOUS COMPOUND POLLUTING OUR AIR. The initial state is n = 1 (vibrations) and l = 0 (rotations) is the initial state. Remember the selection rules are change in l = +1 or -1 and (change in n) = +1 or -1. In the case of photon emission, we know the molecule has (change in n) = -1, which means it drops into the n = 0 state. In that case, l = 1 for the final state. See figure 42.8 which shows an n = 2, l = 0 state dropping into an n = 1, l =1 state for a nice analogy our case of an n = 1, l = 0 state dropping into an n = 0, l = 1state.

If you see the absorption spectra for rotational molecules, we see the molecules can either gain or lose one unit of angular momentum l as well. Thus, logically, the emission spectrum would also have a gain or loss of one unit of angular momentum.

Thus, the final state must be n = 0 and l = 1. The initial energy ( n = 1, l = 0) is
E_i = (3/2)*h_bar*w ands the final state (n = 0 and l = 1) has
E_f= (1/2)*h_bar*w + 1*(1 + 1) (h_bar²)/2I = (1/2)*h_bar*w + 2*(h_bar²)/2I =
(1/2)*h_bar*w + h_bar²/I.

Evaluate E_i- E_f= h_bar*w - h_bar² /I. MANY STUDENTS NEGLECTED THE SECOND NEGATIVE TERM, WHICH WILL CHANGE YOUR ANSWER BY A NEGLIGIBLE FRACTION, WHICH YOU SHOULD HAVE DEMONSTRATED FOR FULL CREDIT.

E_i- E_f= h_bar*w - h_bar² /I = 0.2690 eV - h_bar² /I = 0.2690 eV - 0.000479 eV which still is approximately 0.2690 eV; you should include both vibrational and rotational terms and show the rotational energy change is much, much smaller than the vibrational energy change to justify neglecting the former. See example 42.2 for explicit computation of the rotational energy term; meanwhile it is easy to show h_bar*w = 0.2690 eV as example 42.3 explicitly states, from which you can get k' = 1902 N/m and then work backwards to get w and hence h_bar*w = 0.2690 eV. Now set

hf = 0.2670 eV , from which we get f = 6.5 x10 ¹³ s^-1(Hz), using h = 4.136x10 ^-15 eV-s.

NOTE: Photon frequency f is NOT RELATED to w via f = w/(2*pi)

9. SIMPLE COMPUTATIONS AND BUILDING INTUITION WITH THE FERMI -DIRAC DISTRIBUTION: For a solid metal,

(a) (10 points) what is the probability at room temperature (T = 300 K) for an electron to have energy   0.10 eV above the Fermi energy? In other words, what is the probability that a state with this energy is occupied?
(b) (10 points)   what is the probability at room temperature   for an electron to have energy   0.10 eV below Fermi energy? In other words, what is the probability at room temperature that a state with this energy is occupied?
(c) (5 points) what is the probability (at room temperature)   that the state in part (b) is unoccupied. Use common sense/intuition. The probability of the state being either occupied or unoccupied must be what number? That is, the sum of the probability of the state being occupied and the probability of it being unoccupied must be what number?
(d) (5 points) Compute the energy E of a conduction electron in silver at 800 K if the probability of finding an electron in that state is 0.950. Assume the Fermi energy is 5.48 eV at this temperature.

NOW SUPPOSE THE METAL'S TEMPERATURE IS REDUCED TO ABSOLUTE ZERO:

(e) (2 points) What is the probability at absolute zero (T = 0 K) for an electron to have energy   0.10 eV above the Fermi energy?
(f) (2 points) What is the probability at absolute zero (T = 0 K) for an electron to have energy   0.10 eV below the Fermi energy?
(g) (5 points) At T = 0K, the Fermi level of copper is 7.03 eV. What is the speed of an electron that has an energy equal to the average energy at T = 0K?

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SOLUTIONS:
See figures 42.24 and 42.25 for this problem.

(a) f(E) = 1/(e^(E-EF^)/kT   + 1) = 1/(e^0.10/kT   + 1) . Now, 0.1 eV/kT =
0.16x10 ^-19J/[1.38x10 ^-23*300 J] = 3.86 . Thus, f(E) = 1/(e^3.86   + 1) = 0.0206 = 2.06 %.

(b) f(E) = 1/(e^(E-EF^)/kT   + 1) = 1/(e^-0.10/kT   + 1) . Now, 0.1 eV/kT =
-0.16x10 ^-19J/[1.38x10 ^-23*300 J] = -3.86 . Thus, f(E) = 1/(e-^3.86 + 1) = 0.9794 =
97.94 %.

(c) 1- 0.9794 = 0.0206 = 2.06 %

(d) SEE EXAMPLE 42.7: IF f(E) = 1/(e^(E-EF^)/kT + 1) = 0.950 , THEN,

E = E_F + kT*ln[1/f(E) - 1] = 5.48 eV + (8.625 x10^-5 eV/K)(800 K)*ln[1/0.950 - 1]

= 5.48 eV + (0.025875)*ln(1.0526 - 1) = 5.48 + (0.069)*(-2.94444) = 5.48 - 0.076187 = 5.28 eV.

(e) 0; see figure 42.24 and 42.25.

(f) 1; see figure 42.24 and 42.25.

(g) E_av = (3/5)*E_F, where E_F is the Fermi energy at T = 0, when the system is in a thermal "ground state." Thus, (1/2)*mv_av² = (3/5)*E_{F ,}which gives v_av = 1.2x10 ⁶ m/s.

10. ENERGY BANDS AND SEMI-CONDUCTORS:
(10 POINTS) The gap between the valance and conduction bands in Silicon is 1.12 eV. A radioactively "hot" nickel nucleus in an excited state emits a high energy gamma ray photon with a wavelength 9.00x10^-4 nm. How many electrons can be excited from the top of the valence band to the bottom of the conduction band by the absorption of this gamma ray?
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SOLUTIONS:
hc/lambda = Energy = n*1.12 eV, where n is the number of electrons excited into the valance band; for the given wavelength lambda, we get n = 1.24x10 ⁶electrons

11. SEMI-CONDUCTOR DEVICES. Last time we reviewed how the a p-n junction is characterized by a space charge region, the result of the diffusion of electrons from the n side to the p side. Visit:
http://en.wikipedia.org/wiki/P%E2%80%93n_junction .
This diffusion ceases when an opposing electric force cancels out the "diffusion " force. That force is caused by an electric field from the two adjacent oppositely charged layers, one negative (on the p side, from the build up of electrons) and the other positive (on the n side, from the removal of electrons). In order for positive current to flow into the p-side, the net electric field must be reduced by an oppositely directed field set up by an external battery whose positive terminal is connected to the p-side in what we call "forward bias."
(a) (10 points) A forward bias voltage of 15.05 mV produces a positive current of 9.24 mA through a p-n junction at 300.00 K. What is the positive current to become if the forward-bias voltage is reduced to 10.00 mV?
(b) (5 points) In reverse bias, the battery's negative terminal is connected to the p-side. For reverse bias voltages of -15.01 mV and -10.02 mV, what is the reverse-bias negative current?

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SOLUTIONS: Use equation I = I_s*(e^eV/kT -1); see figure 42.31, which shows a current rising dramatically in forward bias and very gradually at reverse bias. If we had time, we would compute the "cut-off" voltage when the current begins to rise dramatically with very large slope. That computation is easy when you determine criteria on the slope.
(a) First you need to get factor I_s , know as the saturation current, using information given: I = 9.25 mA when V = 15.05 mV. That gives I_s = 0.0117 (A), in turn giving   I = 5.53 mA when V = 10.00 mV. We reduced the voltage by 33 % and got a reduction in current of some 40 %. It appears we were below the "cut-off " voltage shown in figure 42.31, which shows a much more dramatic variation of current with voltage at larger values of potential.
(b) Now that you know the saturation current I_s, the rest is easy: Plug in the two values of voltage and you a negative current I, consistent with figure 42.31.
I = -5.16 mA when V = -15.01 (mV)
I = -3.75 mA when V = -10.02 (mV)
We reduced the voltage by 33 % and got a reduction in current of nearly 38 % . In figure 42.31, going in the negative potential direction it appears we are in the region of the graph before the curve flattens out at larger magnitude negative voltages.