Physics 232 Exam III
Apr. 28, 2006
Soc. Sec #
Name
1. A photosensor detects a total of 3 × 1015 photons in one second. The photosensor also
measures in one second a total energy of 35 joules. What is the wavelength of these
photons?
The energy of one photon is given by the total energy divided by the total number of photons
Eγ
=
=
ET otal
Nγ
35
= 1.1667 × 10−14 joules
3 × 1015
The wavelength of an individual photon is related to its energy by
Eγ
hc
¢¡
¢
¡λ
6.626 × 10−34 3 × 108
hc
=
Eγ
1.1667 × 10−14
= hf =
λ =
= 1.704 × 10−11 meters
1
Soc. Sec #
Name
2. Light of wavelength 450nm strikes a surface causing the ejection of photoelectrons for
which the stopping potential is given by Vstop = 0.35 V olts.
This is a problem in the photoelectric effect.
(a) What is the work function of the surface?
The photoelectric effect equation is
K.E. = Eγ − Φ
where Φ is the work function. The kinetic energy of the electron is given by the work necessary
to be done against the electron
K.E. = q V
We then have
qV
= Eγ − Φ
hc
Φ = Eγ − q V =
−qV
λ ¢¡
¢
¡
¡
¢
6.626 × 10−34 3 × 108
− 1.6 × 10−19 (0.35)
=
−9
450 × 10
= 4.4173 × 10−19 − 5.6 × 10−20
= 3.8573 × 10−19 joules
= 2.411 eV
(b) What is the threshold frequency for this surface?
The threshold wavelength can be gotten by setting the kinetic energy of the electron to zero.
K.E. = Eγ − Φ = 0
Eγ = Φ
hf = Φ
Φ
3.8573 × 10−19
f =
= 5.821 × 1014 Hz
=
h
6.626 × 10−34
(c) Now suppose radiation of wavelength 200nm is incident on the surface. What is now
the necessary stopping potential for the emitted photolectrons?
K.E. = Eγ − Φ
q V = Eγ − Φ
=
V
=
¢¡
¢
¡
6.626 × 10−34 3 × 108
hc
− 3.8573 × 10−19
−Φ=
λ
200 × 10−9
9.939 × 10−19 − 3.8573 × 10−19
= 3.80 V olts
1.6 × 10−19
2
Soc. Sec #
3.
28
13 Al
Name
has a half-life of 2.24 minutes. Initially it has decay rate of 200 decays/minute.
(a) What is the decay constant for this isotope?
The decay constant is related to the half-life by
t1/2
=
λ =
0.693
0.693
or λ =
λ
t1/2
0.693
0.693
=
2.24 (60)
134.4
= 0.005156 seconds−1
(b) What percentage of this isotope will be left after one hour?
The number of isotopes that is left is given by
N
N
N0
P ercentage
= N0 e−λ t
= e−λ t = e−(0.005156)(3600) = e−18.5616 = 8.686 × 10− 9
= 100 ×
N
= 8.686 × 10− 7
N0
(c) How long from the initial time will it take for the sample to reach an activity of 50.0
decays/minute?
The change in activity is given by
R0 e−λ t
200 e−(0.005156)t
e−(0.005156)t
− (0.005156) t
− (0.005156) t
1.3863
t =
= 268.87 seconds = 4.48 minutes
0.005156
R
50
0.25
ln(0.25)
−1.3863
=
=
=
=
=
3
Soc. Sec #
Name
4. Imagine that the only visible photons that an electron in a box, of infinite depth and of
unknown length, emits have wavelengths of 689 nm and 413 nm. Identify the transitions
and find the length of the box.
For a particle in a box, the energy levels are given by
En =
n2 π 2 h̄2
2 m L2
In general the energy released in a transition is given by
∆E =
¢
hc
π 2 h̄2 ¡ 2
ni − n2f
=
2
λ
2mL
Let λ1 = 689 nm and λ2 = 413 nm. We have
∆E1
=
∆E2
=
hc
π2 h̄2
=
λ1
2 m L2
hc
π2 h̄2
=
λ2
2 m L2
We take the ratio of equations 1 and 2 to get
¡ 2
¢
n1i − n21f
¡ 2
¢
n2i − n22f
n21i − n21f
λ2
413
= 2
=
= 0.5994
2
λ1
n2i − n2f
689
(1)
(2)
(3)
As all the n values are integers, the right hand side of equation 3 should be that of the ratio of integers.
We therefore set the right hand side to 0.6 which is the ratio of 3 to 5
n21i − n21f
3
λ2
= 2
= 0.6 =
λ1
n2i − n22f
5
We then have that
n21i − n21f = 3
(4)
The only combination of integers that satisfies equation 4 are
n1i = 2 and n1f = 1
and
n22i − n22f = 5
The only combinaion of integers that satisfies equation 5 are
n2i = 3 and n2f = 2
Now for the length of the box. We can use either equation 1 or equation 2. We will use 1
∆E1
=
L2
=
=
¢
hc
π2 h̄2 ¡ 2
n1i − n21f
=
2
λ1
2mL
¢ 3λ1 h
λ1 π 2 h̄2 ¡ 2
n1i − n21f =
2¡m h c
8mc ¢
¢¡
3 689 × 10−9 6.626 × 10−34
= 6.271 × 10−19
8 (9.1 × 10−31 ) (3 × 108 )
L = 7.92 × 10−10 meters
4
(5)
Physics 232 Formula Sheet
Simple Harmonic
Motion
Waves on a String
r
r
F = −k x
x = A cos(ω t + φ )
v=
T
λ=
2L
n
k
m
ω=
µ
n = integer
ω = 2π f
1
T=
f
Reflection
1
k x2
2
E Total = PE + KE
Refraction
E Total
sin θ c =
na
nb
na < nb
I = I 0 cos 2 φ
Electromagnetic
Waves
Periodic Motion
y( x, t ) = A cos(k x ± ω t )
y( x, t ) = 2 A cos(k x) sin(ω t )
∂ 2 y( x, t )
∂x 2
v =λ f
=
1 ∂ 2 y( x, t )
v2
v'x =
v' y =
Polarization
v + vL
fS
v + vS
∂t 2
ynet = ∑ yi
i
1
Pave = µ v ω 2 A2
2
t − v x / c2
na sin θ a = nb sin θ b
Doppler Effect
fL =
1 − v2 / c2
y' = y
z' = z
t' =
θ incident = θ reflected
1
= k A2
2
x−vt
x' =
Light
PE =
Special Relativity
Emax = c Bmax
1
I = Save = ε 0 c E 2
2
r
1 r r
S=
E×B
µ0
v 'z =
T=
1 − v2 / c2
vx − v
1 − (v / c 2 ) v x
v y 1− v2 / c2
1 − (v / c 2 ) v x
vz 1 − v 2 / c 2
1 − (v / c 2 ) v x
T0
1 − (v c )2
L = L0 1 − (v c )2
E=
m0c 2
1 − (v c )2
Diffraction
E = KE + m0 c 2
d sin θ = m λ
E 2 = p 2 c 2 + m02 c 4
sin θ1 = 1.22
λ
D
Refraction at a Curved
Surface
n n' n'− n
+ =
s s'
R
n s'
m=−
n' s
Interference
DeBroglie Hypothesis
δ = δ inh + δ p.d . + δ refl
δ p.d . = 2π
∆x
λ
Two Slit Intensity
⎛π d
⎞
sin θ ⎟
I = I 0 cos 2 ⎜
⎝ λ
⎠
Reflected Intensity
Mirrors
1 1 1
+ =
s s' f
R
s'
f = ;m =−
s
2
Photoelectric Effect
Lenses
KE = h f − B .E .
I reflected
1 1 1
+ =
s s' f
s'
m=−
s
λmaxT = 2.90 × 10 − 3 m K
ITotal = σ T 4
2π h c 2
h
(1 − cos θ )
me c
)
h
p
Heisenberg
Uncertainty Principal
h
2π
h
∆E ∆ t ≥
2π
∆p x ∆ x ≥
Particle in Infinite Well
En =
n 2π 2 h 2
kn =
En = −
rn = ε 0
1 me e 4
ε 02 8 n 2 h2
13.6
n2
eV
n 2 h2
π me e 2
rn = n 2 (0.53 Angstroms )
2 m L2
2mE n
h
ψ n ( x) =
Bohr Model
En = −
λ5 e h c / k T − 1
E = nh f
λ '−λ0 =
h
L = mvr = n
2π
Blackbody Radiation
(
2
Compton Scattering
⎛ 1
1
1 ⎞
⎟⎟
= (n − 1)⎜⎜
−
f
⎝ R1 R2 ⎠
I (λ ) =
⎛ n − nb ⎞
⎟⎟
= ⎜⎜ a
+
n
n
⎝ a
b⎠
λ=
2
⎛ nπ x ⎞
sin⎜
⎟
L ⎝ L ⎠
Hydrogen Atom
En = −
1 me e 4
ε 02 8n 2 h 2
n = 1, 2, 3, K
L = l (l + 1) h
l = 0, 1, 2, K , n − 1
Lz = m h
m = 0, ± 1, ± 2, K , ± l
Physical Constants
Nuclear Stuff
c = 3 × 10 8 m / s
v sound = 1100 ft / sec
E B = Z M H + N M N − ZAM c 2
ε 0 = 8.854 × 10
−12
C / N ⋅m
2
µ 0 = 4π × 10 −7 T m / A
e = 1.6 × 10 −19 C
m e = 9.109 × 10 −31 kg
= 0.511 MeV / c 2
m p = 1.672 × 10 − 27 kg
= 938.27 MeV / c 2
1eV = 1.6 × 10 −19 joules
W
σ = 5.67 × 10 −8 2 4
m K
h = 6.626 × 10 −34 joule sec
Useful Geometry
Circle
Area = π r 2
Circumfere nce = 2π r
Sphere
Surface Area = 4 π r 2
4
Volume = π r 3
3
Cylinder
Lateral Area = 2 π r L
Volume = π r 2 L
(
2
)
N = N 0 e −λ t
R=−
t1 =
2
τ=
1
λ
dN
= R0 e − λ t
dt
0.693
λ