Phys 274 c Eric B. Szarmes Weekly Homework 004 SOLUTIONS Problem 1 Q5T.1, page 92.
14 February 2011
(a) When light goes through a slit, the beam broadens somewhat, and if it is projected on a screen,
Phys 274 c Eric B. Szarmes Daily Homework 8 SOLUTIONS Q4S.5 a) Helium atoms with a deBroglie wavelength of 0.103 nm have a momentum of p = mv = h 6.626(1034 ) J s = = 6.43(1024 ) kg m/s . 0.103(109 )
Phys 274 c Eric B. Szarmes Weekly Homework 003 SOLUTIONS Problem 1 Single-slit diraction
7 February 2011
The given integral is straightforward to evaluate. For integration over the slit coordinate z w
Phys 274 c Eric B. Szarmes Weekly Homework 002 SOLUTIONS Problem 1 Q2S.12
31 January 2011
If Legolas can discern hair color between two people far away, then he can resolve distances down to about the
2011 E. Szarmes
PHYS 274 Q1R.2 a)
WEEKLY ASSIGNMENT 1 DUE MONDAY, JAN 24/11
PROF. SZARMES
Although buzzing lips do not emit a single frequency, there is a vaguely discernable pitch which emphasizes v
Phys 274 Eric B. Szarmes Daily Homework 18 SOLUTIONS Q8B.6
4 March 2011
Consider the possible transitions in the box potential, which I have arranged below in order of increasing energy dierence (i.e.
Phys 274 c Eric B. Szarmes Daily Homework 17 SOLUTIONS Q8B.2
2 March 2011
The longest wavelength emitted by an electron in a box corresponds to the 2 1 transition (ninit = 2; nnal = 1), which has the
Phys 274 c Eric B. Szarmes Daily Homework 16 SOLUTIONS Q7S.1
28 February 2011
For a quanton in an innite square-well (i.e. box) potential of length L, we have the fundamental quantization condition th
Phys 274 c Eric B. Szarmes Daily Homework 15 SOLUTIONS Q7B.2
25 February 2011
For an electron conned to a wire, which we consider to be a one-dimensional box of length L, the energy of the nth energy
Phys 274 c Eric B. Szarmes Daily Homework 14 SOLUTIONS Q6S.8
18 February 2011
Graphs of the wavefunction and its square are shown below. The wavefunction is (x) = A sin(3x/L) for 0 x L, and (x) = 0 fo
2011 E. Szarmes
PHYS 274 Q6S.2
DAILY ASSIGNMENT 13 DUE WEDNESDAY, FEB 16/11
PROF. SZARMES
According to the superposition principle, the electrons in the recombined beam remain in the |+z spin state,
Phys 274 c Eric B. Szarmes Daily Homework 12 SOLUTIONS Problem 1 Q6T.2 ; Q6T.3 Q6T.2
14 February 2011
A: The probability is 0, because the state | = 1/ 2 , 1/ 2 is the | + x state. Therefore, all of
Phys 274 c Eric B. Szarmes Daily Homework 11 SOLUTIONS Q5T.2 The table is lled out with the corresponding patterns as follows: Case Sample (a) (b) (c) (d) (e) Here are the steps: Wavelength 5 nm 5 nm
Phys 274 c Eric B. Szarmes Daily Homework 10 SOLUTIONS Q5S.2
9 February 2011
We are asked to consider the following series of Stern-Gerlach devices (the y and z directions are drawn perpendicular to t
Phys 274 c Eric B. Szarmes Daily Homework 9 SOLUTIONS Q5B.5 We are asked to consider the following series of Stern-Gerlach devices:
z y z y y ! z
7 February 2011
2 P(+!) = cos ! 2
+
+
?
SG(z)
z y
SG(!
Phys 274 c Eric B. Szarmes Daily Homework 7 SOLUTIONS Q4B.1
2 February 2011
The electrons in this problem have a kinetic energy of K = 25 eV, and thus are non-relativistic since K mc2 , where mc2 = 51
Phys 274 c Eric B. Szarmes Daily Homework 6 SOLUTIONS Q3S.7
31 January 2011
The maximum potential Vmax developed between the plates is related to the maximum kinetic energy of the ejected electrons by
2011 E. Szarmes
PHYS 2 7 4
DAILY A SSIGNMENT 5 DUE FRIDAY, JAN 28/11
PROF. SZARMES
Q3B.2
A photon energy of 3.5 eV corresponds to E = 3.5 eV = hc , where the fundamen tal constant h c = ! 1240 eVnm.
2011 E. Szarmes
PHYS 274
DAILY ASSIGNMENT 4 DUE WEDNESDAY, JAN 26/11
PROF. SZARMES
Q2S.2 D = 5.0 km f = 100 MHz = c/f = 3(108)m/s/100(106) Hz = 3.0 m
Note that radio waves travel at the speed of ligh
2011 E. Szarmes
PHYS 274 Q2B.2
DAILY ASSIGNMENT 3 DUE MONDAY, JAN 24/11
PROF. SZARMES
According to Eq. Q2.1, the angles of constructive interference n from two sources depend only on the wavelength a
2011 E. Szarmes
PHYS 274 Q1S.2
DAILY ASSIGNMENT 2 DUE FRIDAY, JAN 21/11
PROF. SZARMES
For the fundamental mode with two ends fixed, we have
f1 = v (1) = 1 2L 2L
where L = 25 in = 0.635 m and = 0.2 g/
2011 E. Szarmes
PHYS 274 Q1B.1
DAILY ASSIGNMENT 1 DUE WEDNESDAY, JAN 19/11
PROF. SZARMES
The following figures represent the positions of the waves (each moving with a speed of 5 cm/s) at times t = 0