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 ) m
4 February 2011
(1)
Since the mass of a helium atom i
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, one can see brights and dark fringes on the slit image.
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 we obtain
a/2
ys (x) =
a/2
cos(kx kz ) dz
a/2 1 sin(kx k
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 width of a head (so that he can discern a yellow blob
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 vibrations within a narrow range of frequencies. If this
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. increasing photon energy ): ni 2 3 4 3 5 nf 1 2 3 1 4
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 smallest energy spacing and hence the longest wavelengt
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 that 2L n = ; n = 1, 2, 3, . (1) n (The perfectly reectin
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 eigenstate is (hc)2 n2 h2 n2 = . (1) En = 8mL2 8 mc2 L2
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 for x < 0 and x > L.
!(x)
A 0 L
A2
|!(x)| 2
x
0
1 3L
2 3L
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, but are expressed as a superposition of the |+ and | st
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 the electrons will emerge from the (+) channel of the S
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 5 nm 10 nm 10 nm 10 nm Slit Width a 3 6 6 6 6 6 m m m m
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 the beam direction at the indicated points):
z y z y
#
+
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(!)
! y z
2 P( !) = sin ! 2
I have also indicated schemat
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 = 511,000 eV for electrons. We therefore may use = hc 2Kmc2
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 the relation Kmax = eVmax . Thus, if ultraviolet light
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. Thus, ! = hc = 1240 eVnm = 354 nm [ultraviolet, not vis
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 light, not the speed of sound! Now, for two antennas emitti
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 and the slit separation d between the sources (and not t
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/m = 0.0002 kg/m. a)
F
If f1 = 329 Hz, then we require
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, 2, 3, 4, and 6 s. The dashed lines show their individ