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Mid Term Exam 1
Phys 248
Feb 13, 2009
Print your name and ID number clearly above.
To receive full credit you must show all your work. If
you only provide your final answer (in the boxes)
and do not show your work you will only receive
partial credit, even for a correct final answer.
Problem
Score
1
/20
2
/20
3
/20
4
/30
Total
/90
Useful constants:
Planck constant h = 6.626 x 10-34 J s
Reduced Planck constant = h = 1.054 x 10-34 J s
hc = 1240 eV nm
h c = 197.3 eV nm
c = 3 x 108 m/s
1 eV = 1.6 x 10-19 J
! = 9.11 x 10-31 kg = 0.511 MeV
me = electron mass
!
Problem 1: short problems.
1.
A bat can detect very small objects, such as insects whose
length is approximately equal to one wavelength of the sound the
bat makes. If the bat emits chirps at a frequency of 55.0 kHz,
and the speed of sound in air is 340 m/s, what is the smallest
insect the bat can detect?
d
d="=
!
Units
v
340
=
= 6.18mm
f 55 #10 3
2. A particle executes simple harmonic motion with an
amplitude of 4.00 cm. At what position does its speed equal half
its maximum speed?
v max = "A
"A
v=
2
1 2 1 2 1 2
" 2 A2
2
2
2 2
E = E k + U # kA = mv + kx # " A = v + " x =
+ "2x2 #
2
2
2
4
3
3
x 2 = A2 # x = ±
A = 3.46cm
4
2
!
x
Units
3.
A seismic wave far from the epicenter of an earthquake can
be modeled as a wave on a string transporting energy (we
neglect any absorption). The wave moves from granite into mud
fill with same density but lower bulk modulus. The speed of the
wave drops by a factor of 25.0 with negligible reflection of the
wave. Will the amplitude of the ground shaking increase or
decrease? By what factor?
The rate of energy transfer is:
P = const vA2 => vgraniteAgranite2 = vmudfill Amudfill2 =>
Amudfill = Agranite
v granite
= 5Agranite
v mudfill
Amud fill = ______ x Agranite
Increase
!
4. Standing wave vibrations are set up in a crystal goblet with 4
nodes and 4 antinodes around the 20.0 cm circumference of its
rim. If transverse waves move around the goblet at 450.0 m/s
what would be the high harmonic frequency an opera singer has
to produce to shatter the goblet with a resonant vibration?
" 20
=
= 5.0cm # " = 10.0cm
2 4
v
450
f = =
= 4.5kHz
" 10 $10%2
dNN =
f
Units
!
Problem 2: multiple choice. Circle your answers.
1.
The graph shows the kinetic energy of photoelectrons
ejected from a certain metal as a function of the frequency of the
incident light on the metal. What is the work function of the
metal?
(a) 50 eV (b) 2 eV (c) 5.0 x 10-20 J (d) 0 eV (e) 33 x 10-19 eV
(b)
The work function is the point where the line K max = hf " # intercepts
the y axis, that is Kmax = -Φ. From the graph the line has to go
through
!
(x,y) = (5.0x1014Hz, 0)
0 = hf- Φ
0 = 6.626 "10#34 " 5.0 "1014 # $ % $ = 33.13 "10#20 J = 2.07eV
!
2.
The circular pupil of the human eye has a diameter of 4.0
mm. The minimum light intensity that the eye is sensitive to is
1.0 x 10-10 W/m2. Approximately how many photons per second
of wavelength 550 nm does this correspond to?
A) 350 s-1 B) 35000 s-1 C) 3.5 x 1012 s-1 D) 3500 s-1
E) 3.5 x 10-2 s-1
# photons
Power " Area
Power " Area Power " Area " #
=
=
=
=
second
Energy / photon
hf
hc
=
1"10$10 " % " (4.0 "10$3 /2) 2 " 550
= 3483.6s$1 ' 3500s$1
$19
1240eV & nm "1.6 "10
D)
!
3. Find the de Broglie wavelength of a beam of electrons with
kinetic energy 100 eV.
A) 0.710 nm B) 0.130 nm C) 0.180 nm D) 0.122 nm E) 0.550
nm
Electrons with Ek = 100 eV are not relativistic since Ek << mc2.
2mc 2 E k
h
1240eV $ nm
p=
"#= =
= 0.122nm
c
p
2 % 0.511%10 6 %100
D)
!
4.
The beam of electrons in #3 is used in a Davisson-Germer
experiment. The beam travels toward a crystal of inter-planar
spacing d = 3.5 Å. Compute the angle θ where there is a strong
reflected peak that is equal to the incidence angle (see figure).
d
(a) 50o (b) 80o (c) 45o (d) 60o (e) 65o
According to the figure the path difference between the two
beams is: 2d cos "
!
Constructive interference:
%$(
o
o
* = 79.96 + 80 .
& 2d )
#1
2d cos " = m# $ for m=1 " = cos '
!
(b)
Problem 3:
single slit diffraction.
!
A diffraction pattern is formed on a screen 120 cm away from a
0.400 mm wide slit. Monochromatic 546.1 nm light is used.
a) What is the width of the central maximum?
b) Calculate the fractional intensity I/Imax at a point on the screen
4.10 mm from the center of the principal maximum.
a)
Width
tan ! min =
"y
$
L$
120cm & 546.1 & 10 '7 cm
# sin ! min = % 2"y = 2
=2
= 0.328cm
L
a
a
0.04cm
b) The distance Δy corresponds to the angle:
sin " # tan " =
!
$y
L
Units
The total phase difference between waves from the top and
bottom of the slit is:
"=
2#
asin %
$
and the intensity at a point on the screen divided by the
maximum intensity corresponding to θ=0 is
2
!
I /Imax
2
2
#sin(" /2) & #sin()asin * / +) & #sin()a,y /(L+)) &
.2
=%
( =%
( = 1.62 -10
( =%
$ " /2 ' $ ()asin * / +) ' $ ()a,y /(L+)) '
I/Imax
Units
!
Problem 4: wave function and probability.
The wave function of an electron in a box is
$ 3#x '
"(x) = Asin&
)
% L (
!
for 0 ≤ x ≤ L and zero elsewhere, where A is a normalization
constant and L = 10 nm.
a) If we perform an experiment to measure the location of the
electron, what is the probability that we find it between x=L/3
and x = L/2?
b) If the energy of the electron is measured, what is its value?
c) Estimate the energy of the fundamental state by means of the
uncertainty principle.
The probability to find the particle between 0 and L is 1, hence
we determine A:
L
1=
L
% 3$x (
L
*
2
"
"
dx
=
A
#
# sin2'& L *)dx = A 2 3$
0
0
= A2
!
L
3$ , A =
6$
2
L
L
1
# 2 (1+ cos(2y ))dy =
0
Probability
L /2
P=
# "*"dx =
L /3
=
% 6$x ((
2 L / 2 2 % 3$x (
1 L / 2%
sin
dx
=
'
*
**dx =
'1+ cos'
#
#
& L ))
L L /3 & L )
L L / 3&
/ 1
1 ,L L L
+ +
(sin( 3$ ) + sin2$ )1 =
.
0 6
L - 2 3 6$
(3)2 ! 2 h 2 9! 2 h 2
=
b) The particle is in the n=3 state, so E = E3 =
.
2mL2
2mL2
!
Numerically E3 = 3.38e-2 eV
Energy
Units
c) Since the electron is confined in the box:
"x = L # "px $
("px )
!
2
h
L
2
(
= ( px % px ) = px2 % 2 px px + px 2
)
ave
Since the probability that the electron goes in one direction or
the opposite is the same, the average momentum along the
dimension of the box is zero:
2
2
("px ) = px2 # E =
2
px2
1 % h(
1 % h(
,23
,4
$
' * # E min =
' * = 6.1+10 J = 3.8 +10 eV
2m 2m & L )
2m & L )
Emin
!
Units