QUIZ 5 CH. 38: 
Photoelectric effect-1, 4, 9, 10, 12, 13, 56;
Atomic Line Spectra and Energy Levels-18; 
Bohr Model-23, 25, 26, 27; 
Lasers-29, 30;
X-Ray Production and Scattering-34, 36
Not covered in class but please read section 38.7, a straight-forward extension of the
photo-electric effect at higher energies and frequencies)
Continuous Spectra (aka Blackbody radiation)-46, 47
CLICK HYPERPHYSICS  FOR SUMMARIES OF  ALL UNDERGRADUATE  PHYSICS TOPICS  INCLUDING RELATIVITY, ATOMS, NUCLEI AND COSMOLOGY.
SAMPLE EXAMS  #2
CHECK BACK LATER FOR HELPFUL DISCUSSIONS;
We reviewed already # 1, 5, 10, 12, 13, 56 in class--- see lecture
xeroxed (photocopied) notes .  Go to www.nvaphysics.com, i.e.,
http://www.nvaphysics.com/ Phy5SP12/quiz_5_p5sp12.htm,  for the full
list; here are more from the complete problem roster:
Note: #9 is similiar to #56, which we went over in class & on notes. 9. (a) f = c/lambda (b) The ratio (75 W) / hf = P/ hf is what you want. (c) Are the expressions in parts (a) and (b) the same? I think not, since the power P = 75 (W) is proportional to the intensity I, i.e, P = I*A . and the intensity P gives the *accumulated* energy (per unit time) in the beam, INDEPENDENT OF THE NUMBER OF ELECTRONS IN THE BEAM. The actual number of electrons/sec is proportional to this quantity divided by the energy per photon. .
18. See figure 38.36. (a) Clearly it should take 20 eV. Please explain. (b) If 18 eV is absorbed by the atom in the n = 1 state, it will go into the n = 4 state. Please explain why with a simple math statement, akin to balancing your check book. From there it can go into several possible states, assuming there are no special *selection rules* as occurs in the ZEEMAN EFFECT, which we covered last Friday. Think of an airplane flight with stop off points. In this case, from n = 4, the atom (airplane) might go to n = 3, n = 2 as stop off points before continuing on. It could take a 'direct flight' to n = 1. Think of all the possibilities---for example if the first stop off point is n = 3, it might stop off again at n = 2 before continuing on to the ground state! (c) If an 8 eV- photon strikes Searsium, there is *not* enough energy to advance to the n = 2 level. Please explain why with a simple math statement, akin to balancing your check book. More discussions, including part (d) will be sent later today...just wanted to give you a jumpstart.

For more info go to www.nvaphysics.com, i.e.,
http://www.nvaphysics.com/ Phy5SP12/quiz_5_p5sp12.htm,  for full
list; more discussions, i.e.,  18 part (d) + #23, 25 and 26, are below.
NOTE ON NOTATION: IN THE BELOW DISCUSSIONS,  x^2 means x squared. i.e.
x^2 = x*x.

18. (d) Any time an electron is emitted after being struck by a a
photon of a certain energy hf, one can say the following: work
function < hf , a result of the formula:
KE = hf -  work function > 0.
hf - work function > 0 means hf > work function, my original assertion.

In the case at hand work function < 5 eV = hf.
I leave it to you to complete the condition by
replacing the ? with a number (in eV) in the following inequality:
? < work function.
What number ( in eV) should replace the ? symbol in the latter inequality?
Re-read part (d) and the answer is yours.You will have the complete
double inequality: min  < work function < max, where we know max = 5
eV.
23.  En = -13,6 eV/n^2 = -1.51 eV.  Solve for n; n will be extremely
close to an odd integer that will remain unnamed until you find
it---round off n to the closest integer, then use L = n*h/(2*pi),
Bohr's major hypothesis,  to find L.
25. See section 41.4 on approximations for  multi-electron atoms for
more discussion of these results. As stated in that section, you
should be able to derive the result in part (a) below by reading
carefully the straight-forward derivation of  Bohr energy levels in
section 38.5
Z = 4
(a) For Be^3+, En = -(Z^2*13.6 eV)/n^2 in general. Set n = 1 to find
the Be^3+ ground state energy E1. Note the factor Z^2 = Z*Z , where Z
= 4.
(b) The ionization energy is the photon energy required to raise the
electron from the ground state (n = 1) to an energy of zero (n =
infinity), which means the electron has escaped, ionizing the atom.
The required energy is E(infinity) - E1 =
0 - [-Z^2*13.6 eV)/n^2], where n = 1 and Z = 4.
(c)  hc/lambda = E2 - E1. Find lambda and compare with 122 nm for Hydrogen.
(d) Read derivations of equations 38.12 and 38.13 and solve the the
following two equations (i) and (ii) simultaneously for rn as done on
pp1323-4:
(i) m*v^2/rn = (Z*k*e^2)/rn^2
(ii) m*v**rn = n*h/(2*pi)
You will get something like equation 38.12, but there will be a factor
of Z in the denominator.
26. (a) E5 - E2 = hc/lambda.
(b) Read p 1412, Ch. 41, on selection rules, which gives the real deal
on atomic transitions in contrast to the naive approximation in this
problem. The photon is a boson, named after celebrated Indian
physicist Bose, who connected with Einstein to come up with
'Bose-Einstein' statistics. A boson is a particle  with spin angular
momentum based on a *whole* integer. (A lepton, in contrast,  has spin
based on a half -integer such as 1/2 in the case of the electron, the
world's most famous lepton!)  Without this post-Bohr historical
background one might naively think the following:  The change  in
angular momentum of the atom has magnitude
5*h/(2*pi) - 2*h/(2*pi)  = 3*h/(2*pi); the atom loses angular
momentum, so the change is -  3*h/(2*pi). If angular momentum is
conserved, the emitted photon would have angular momentum of the same
magnitude as the *change* in angular momentum  experienced by the
atom.

As noted in Ch. 41, the modern quantum picture gives a different
result: The photon ordinarily carries off *one* unit (h/[2*pi]) of
angular momentum, which leads to the requirements that in a transition
the quantum number L must change by 1 and mL must change by 0 or +1 or
-1---read page 1412, on selection rules, within the coverage of last
lecture (FRI 4/6).
This is installment 3 of quiz 5 notes.

For more info go to www.nvaphysics.com, i.e.,
http://www.nvaphysics.com/ Phy5SP12/quiz_5_p5sp12.htm,  for full
list; more discussions below
THIS INSTALLMENT~
Bohr Model-27;
Lasers-29, 30;
NEXT INSTALLMENT~
X-Ray Production and Scattering-34, 36
Not covered in class but section 38.7, a straight-forward extension of
photo-electric effect at high energies.
Continuous Spectra (aka Blackbody radiation--. -46, 47 A  good
understanding could come from 4C, which covers heat transfer---
conduction, convection, and radiation; radiative heat transfer  theory
employs the black body.  In 4C, you've  enough to  master Quiz 5.
Subject could occur in multiple choice section of exam. All you need
to understand are quiz problems addressing subject, any exam  problem
will likely be very similar or in multiple choice.

NOTE ON NOTATION: IN THE BELOW DISCUSSIONS,  x^2 means x squared. i.e.
x^2 = x*x.

27. (a) See installment 2, # 26, which relies on eqns 38.12, 13.(b)
(2*pi*Rn)/Tn = Vn , where Rn is the quantum radius and Vn is the
quantum speed indexed by n. I use caps R amd V only to get subscript
n. Evaluate for n = 1, 2 , 3. (c) Set n = 2 to perform calculation of
number of orbits from ratio speed over circumference.
The next two were covered in installment 1 (# 9 ) and in most recent
in-class notes  distributed a week after lecture. In 29 you are
computing a ratio; #30 contains elements within 29.
30. I'll explain (c) only since (a) and (b) are in #9.  c*t = length
of cylindrical pulse.You want to find the number of photons in the
volume of the pulse.  It can be found in this way: If P is the given
power, then  P/hf = # photons per sec passing through an area equal to
cross-sectional area of cylinder. Simply multiply this ratio by time
t.
For more info go to www.nvaphysics.com, i.e.,
http://www.nvaphysics.com/ Phy5SP12/quiz_5_p5sp12.htm,  for full
list; more discussions below
X-Ray Production and Scattering-34, 36
Not covered in class but section 38.7, a straight-forward extension of
photo-electric effect at high energies.
Continuous Spectra (aka Blackbody radiation--. -46, 47 A  good
understanding could come from 4C, which covers heat transfer---
conduction, convection, and radiation; radiative heat transfer  theory
employs the black body.  In 4C, you've  enough to  master Quiz 5.
Subject could occur in multiple choice section of exam. All you need
to understand are quiz problems addressing subject, any exam  problem
will likely be very similar or in multiple choice.
Meanwhile, see sample exams for  ties to active quiz problems.
http://www.nvaphysics.com/ ImSamT4_4C/t4fau01.htm--- problem 2 and
http://www.nvaphysics.com/ ImSamFinal_4C/fsp01.htm--- problem 2; early
next week.we will supplement with chronologically ordered past exams,
which only sample possibilities; be prepared for different questions,
same difficulty.
TEST 2 COVERS QUIZ 4, 5 AND 6. As far as Quiz 6 goes, only  the first
6 problems are required and in this situation , when the due date is
aligned with test date and little time for review,. we make the test
very close to  quiz problems  to check if you are keeping up, we
repeat quiz 6 on test 3 (TBA).
34.(a) charge*potential =  h*c/lambda, where lambda is the minimum
wavelength. This corresponds to the highest energy photon emitted when
the electron comes to a dead stop,  losing all its kinetic energy to a
single light particle with  energy h*c/lambda.
(b) charge*potential =  h*c/lambda.
 
36.
(a) See #34 to get the original wavelength.
(b) In the following, lambda is the wavelength of the photon produced
in part (a). lambda' is the longer photon wavelength after it is
scattered.  lambda '  - lambda = (h/mc)*( 1 - cos theta), where theta
is the angle  the photon makes with its original direction of motion.
Note:  lambda ' > lambda, as covered in class;  the right hand side of
the first equation above is never negative  with  a minimum of zero
when theta = 0 (zero scattering of photon, which travels un-deflected
) and a maximum of 2h/mc.when theta = 180 degrees---the electron is
back scattered in a direction opposite the original direction. In the
latter case the photon loses the most energy to the electron.
(c) Transform answer to (b) into eV.
 
46.
(a) Intensity*area = Power = rate of total energy released from body.
 rho * T^4 *area = rho' * T ' ^ 4 *area', where the prime symbol
refers to the hotter object. Use the formula for the surface  area of
a sphere, where the radius of one object is 3 times that of the other.
Assume rho = rho'   = 5.6704 x10 ^ - 8 W/(m^2*K^4), for an ideal black
body  .
(b) Use lambda_m * T = 2.90 x 10 ^ -3 m*K, for an ideal black body

 
47. This is easy up to a point. Differentiate eqn. 38.32 with respect
to lambda and set the derivative equal  to zero. You will be able to
factor out a power of (e^hc/lambda*K*T - 1) to get  eqn. 38.34. After
that, the hard part is getting eqn. 38.33.