Download Chapter 4 Particle Nature of Matter. Solutions of Selected

Chapter 4
Particle Nature of Matter. Solutions
of Selected Problems
4.1
Problem 4.5 (In the text book)
A Thomson-type experiment with relativistic electmns. One of the earliest experiments to
show that p = γme v (rather than p = me v) was that of C. Neumann. The apparatus shown
in Figure (4.1) is identical to Thomson’s except that the source of high-speed electrons is a
radioactive radium source and the magnetic field B is arranged to act on the electron over its
entire trajectory from source to detector. The combined electric and magnetic fields act as a
velocity selector, only passing electrons with speed v, where v = V /Bd (Equation 4.6), while
in the region where there is only a magnetic field the electron moves in a circle of radius r,
with r given by p = Bre. This latter region (E = 0, B = constant) acts as a momentum
selector because electrons with larger momenta have paths with larger radii.
Figure 4.1: The Neumann apparatus.
2
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
(a) Show that the radius of the circle described by the electron is given by r = (`2 + y 2 )/2y.
(b) Typical values for the Neumann experiment were d = 2.51 × 10−4 m, B = 0.0177 T ,
and ` = 0.0247 m. For V = 1060 V , y, the most critical value, was measured to be
0.0024 ± 0.0005 m. Show that these values disagree with the y value calculated from
p = me v but agree with the y value calculated from p = γme v within experimental error.
(Hint: Find v from Equation 4.6, use me v = Bre or γme v = Bre to find r, and use r to
find y.)
Solution
Figure 4.2: Details of electron path in region where E = 0 and B = constant.
(a) Figure (4.2) shows the path of the electron in the region where E = 0 and B = constant.
The path intersect the detector (photo plate) at point f . Using the right-angle triangle
cf d we get:
r2 =
=
2ry =
=
`2 + (r − y)2
`2 + r2 + y 2 − 2ry
−r2 + `2 + r2 + y 2
`2 + y 2
`2 + y 2
r =
2y
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4.1. PROBLEM 4.5 (IN THE TEXT BOOK)
3
(b) The combination of electric and magnetic fields selects a velocity v of the electron, given
by v = V /Bd, using the given numerical values we get:
v =
V
Bd
1060
0.0177 × 2.5 × 10−4
= 2.39 × 108 m/s
=
using the classical expression for the momentum, we get:
p = me v
= Bre
me v
r =
Be
9.11 × 10−31 × 2.39 × 108
=
0.0177 × 1.60 × 10−19
= 7.69 × 10−2 m
Using y 2 − 2yr + `2 = 0 we can find the value of y that corresponds to the above value
of r:
0 = y 2 − 2ry + `2
= y 2 − 2 × 7.69 × 10−2 y + (2.47 × 10−2 )2
= y 2 − 1.54 × 10−1 y + 6.10 × 10−4
using the quadratic equation solution: y = (−b ±
−1.54 × 10−1 , and c = 6.10 × 10−4 , we get:
y =
=
=
=
=
√
b2 − 4ac)/2a, with a = 1, b =
√
b2 − 4ac
2a
p
+1.54 × 10−1 ± (1.54 × 10−1 )2 − 4 × 6.10 × 10−4
2
−2
−2
7.70 × 10 ± 7.29 × 10
1.499 × 10−1 m
or
4.10 × 10−3 m
4.10 × 10−3 m
−b ±
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
The first solution is rejected because it gives y > r.
p
Using the relativistic expression of momentum p = γmv = Bre, where γ = 1/ 1 − v 2 /c2 ,
we get:
p = γme v
= Bre
γme v
r =
Be
mv
p e
=
Be 1 − v 2 /c2
9.11 × 10−31 × 2.39 × 108
p
=
0.0177 × 1.60−19 1 − (2.39 × 108 /3 × 108 )2
= 1.272 × 10−1 m
The quadratic equation in y for this gives:
0 = y 2 − 2.544 × 10−1 y + 6.10 × 10−4
p
+2.544 × 10−1 ± (2.544 × 10−1 )2 − 4 × 6.10 × 10−4
y =
2
−1
−1
= 1.272 × 10 ± 1.248 × 10
= 2.520 × 10−1 m
or
2.400 × 10−3 m
= 2.400 × 10−3 m
The first solution is rejects because it gives y > r and obviously, the value of y calculated
from the relativistic expression agrees well with the experimental value.
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4.2. PROBLEM 4.6 (IN THE TEXT BOOK)
4.2
5
Problem 4.6 (In the text book)
In a Millikan oil-drop experiment, the condenser plates are spaced 2.00 cm apart, the potential across the plates is 4000 V , the rise or fall distance is 4.00 mm, the density of the oil
droplets is 0.800 g/cm3 , and the viscosity of the air is 1.81 × 10−5 kg · m−1 s−1 . The average
time of fall in the absence of an electric field is 15.9 s. The following different rise times in
seconds are observed when the field is turned on: 36.0, 17.3, 24.0, 11.4, 7.54.
(a) Find the radius and mass of the drop used in this experiment.
(b) Calculate the charge on each drop, and show that charge is quantized by considering
both the size of each charge and the amount of charge gained (lost) when the rise time
changes.
(c) Determine the electronic charge from these data. You may assume that e lies between
1.5 and 2.0 × 10−19 C.
Solution
(a) The drag force D on the oil droplet is given by:
D = 6πaηv
where a is the radius of the droplet, η is the viscosity of the air, and v is the velocity
of the droplet. During the fall the droplet moves under the gravity downward and the
drag force upward. When the two forced become equal, the droplet moves with terminal
speed and we get:
mg = 6πaηv
since the droplet is spherical, then m = (4πa3 /3)ρ, where ρ is the density of the droplet,
ad the last equation becomes:
4 3
πa ρg = 6πaηv
3
or
r
a=
Physics 205:Modern Physics I, Chapter 4
Fall 2004
9ηv
2ρg
Ahmed H. Hussein
6
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
The terminal speed of the droplet in the absence of the the electric field is:
v=
∆y
4 × 10−3
=
= 2.52 × 10−4 m/s
∆t
15.9
the radius of the droplet is then:
r
a =
9ηv
2ρg
r
9 × 1.81 × 10−5 × 2.52 × 10−4
2 × 0.8 × 103 × 9.8
= 1.62 × 10−6 m
=
The mass of the droplet is given by:
m = ρV
4
= ρ πa3
3
4 × 0.8 × 103 × π × (1.62 × 10−6 )3
=
3
= 1.42 × 10−14 kg
(b) The charge on the droplet is given by:
mg v + v 0
E
v
where E is the electric field E = V /d = 4000/0.02 = 2.00 × 105 V /m, v is the downward
terminal velocity in the absence of the electric field v = 2.52 × 10−4 m/s, and v 0 is the
upward terminal speed in the presence of the electric field. Using the times given, the
various v 0 s can be calculates as:
q=
v10 =
v20
4 × 10−3
=
= 2.31 × 10−4 m/s
17.3
v30 =
Physics 205:Modern Physics I, Chapter 4
4 × 10−3
= 1.11 × 10−4 m/s
36.0
4 × 10−3
= 1.67 × 10−4 m/s
24.0
Fall 2004
Ahmed H. Hussein
4.2. PROBLEM 4.6 (IN THE TEXT BOOK)
7
4 × 10−3
=
= 3.51 × 10−4 m/s
11.4
4 × 10−3
0
v5 =
= 5.31 × 10−4 m/s
7.54
v40
Now the charges can be calculated:
q =
q1 =
q2
q3
q4
q5
=
=
=
=
=
=
=
=
=
=
=
=
mg(v + v10 )
Ev
1.42 × 10−14 × 9.8(2.52 × 10−4 + v10 )
2.52 × 10−4 × 2.00 × 105
2.761 × 10−15 × (2.52 × 10−4 + v10 )
2.761 × 10−15 × (2.52 × 10−4 + 1.11 × 10−4 )
10.0 × 10−19 C
2.761 × 10−15 × (2.52 × 10−4 + v20 )
2.761 × 10−15 × (2.52 × 10−4 + 2.31 × 10−4 )
13.4 × 10−19 C
2.761 × 10−15 × (2.52 × 10−4 + 1.67 × 10−4 )
11.6 × 10−19 C
2.761 × 10−15 × (2.52 × 10−4 + 3.51 × 10−4 )
16.7 × 10−19 C
2.761 × 10−15 × (2.52 × 10−4 + 5.31 × 10−4 )
21.6 × 10−19 C
(c) Now, we will try to find an integer n1 such that 1.5 × 10−19 < q/n1 < 2.0−19 :
q1
6
q2
8
q3
7
q4
10
q5
13
= 1.67 × 10−19 C
= 1.68 × 10−19 C
= 1.67 × 10−19 C
= 1.67 × 10−19 C
= 166 × 10−19 C
The amount of charges gained or lost are:
|q1 − q2 | = 3.4 × 10−19 C
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
8
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
|q1 − q3 |
|q1 − q4 |
|q1 − q5 |
|q2 − q3 |
|q2 − q4 |
|q2 − q5 |
|q3 − q4 |
|q3 − q5 |
|q4 − q5 |
=
=
=
=
=
=
=
=
=
1.6 × 10−19 C
6.7 × 10−19 C
11.6 × 10−19 C
1.8 × 10−19 C
3.3 × 10−19 C
8.2 × 10−19 C
5.1 × 10−19 C
10.0 × 10−19 C
4.9 × 10−19 C
Once again we find an integer n2 such that 1.5 × 10−19 < (qi − qj )/n2 < 2.0 × 10−19 :
|q1 − q2 |
2
|q1 − q3 |
1
|q1 − q4 |
4
|q1 − q5 |
7
|q2 − q3 |
1
|q2 − q4 |
2
|q2 − q5 |
5
|q3 − q4 |
3
|q3 − q5 |
6
|q4 − q5 |
3
= 1.70 × 10−19 C
= 1.60 × 10−19 C
= 1.68 × 10−19 C
= 1.66 × 10−19 C
= 1.80 × 10−19 C
= 1.65 × 10−19 C
= 1.64 × 10−19 C
= 1.70 × 10−19 C
= 1.67 × 10−19 C
= 1.63 × 10−19 C
Averaging all values of qi /n1 and (|qi − qj |)/n2 , we get:
qave = 1.67 × 10−19 C
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4.3. PROBLEM 4.8 (IN THE TEXT BOOK)
4.3
9
Problem 4.8 (In the text book)
A parallel beam of α particles with fixed kinetic energy is normally incident on a piece of
gold foil.
(a) If 100 α particles per minute are detected at 20◦ , how many will be counted at 40◦ , 60◦ , 80◦ ,
and 100◦ ?
(b) If the kinetic energy of the incident α particles is doubled, how many scattered α particles
will be observed at 20◦ ?
(c) If the original α particles were incident on a copper foil of the same thickness, how
many scattered α particles would be detected at 20◦ ? Note that ρcu = 8.9 g/cm3 and
ρAu = 19.3 g/cm3 .
Solution
(a) Using Equation (4.16), we get:
∆n(φ) ∝
φ
sin
2
4
and
∆n(φ2 ) = ∆n(φ2 )
sin φ21
!4
sin φ22
so,
sin 10◦
∆n(40 ) = ∆n(20 )
sin 20◦
4
0.174
= 100 ×
0.342
= 6.64 cpm
◦
Physics 205:Modern Physics I, Chapter 4
◦
Fall 2004
4
Ahmed H. Hussein
10
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
sin 10◦
∆n(60 ) = ∆n(20 )
sin 30◦
4
0.174
= 100 ×
0.500
= 1.45 cpm
4
sin 10◦
∆n(80 ) = ∆n(20 )
sin 40◦
4
0.174
= 100 ×
0.643
= 0.533 cpm
4
◦
◦
◦
◦
sin 10◦
∆n(100 ) = ∆n(20 )
sin 50◦
4
0.174
= 100 ×
0.766
= 0.264 cpm
◦
◦
4
(b) From Equation (4.16) we get:
∆n ∝
1
Eα2
so doubling Eα reduces ∆n by a factor of 4, so ∆n at 20◦ = 25 cpm.
(c) From Equation (4.16) we get:
∆n ∝ Z 2 N
2
ZCu
NCu
2
ZAu NAu
where Z is the atomic number of the atom of the foil and N is the number of nuclei per
unit area in the foil.
∆n(Cu) = ∆n(Au)
The number of nuclei per unit area in a foil is given by:
N = Number of nuclei per unit volume × foil thickness
= Number of nuclei per gram × foil density × foil thickness
Avogadro’s Number
=
× foil density × foil thickness
Molecular weight of foil in grams
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4.3. PROBLEM 4.8 (IN THE TEXT BOOK)
11
we then get for each foil:
6.02 × 1023
× 8.9 × tCu
63.54
= 8.43 × 1022 t
6.02 × 1023
=
× 19.3 × tAU
197.0
= 5.90 × 1022 t
NCu =
NAu
The number of α particles scattered from the copper foil at 20◦ is:
2
ZCu
NCu
2
ZAu NAu
(29)2 × 8.43 × 1022
= 100 ×
(79)2 × 5.9 × 1022
= 19.3 cpm
∆n(Cu) = ∆n(Au)
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
12
4.4
Problem 4.31 (In the text book)
(a) Find the frequency of the electron’s orbital motion, fe , around a fixed nucleus of charge
+ Ze by using Equation 4.24 and fe = (v/2πr) to obtain
me k 2 Z 2 e4
fe =
2π~3
1
n3
(b) Show that the frequency of the photon emitted when an electron jumps from an outer
to an inner orbit can be written
fphoton
kZ 2 e2
=
2a◦ h
1
1
− 2
2
nf
ni
me k 2 Z 2 e4
=
2π~3
!
ni + nf
2n2i n2f
!
(ni − nf )
For an electronic transition between adjacent orbits, ni − nf = 1 and
fphoton
me k 2 Z 2 e4
=
2π~3
ni + nf
2n2i n2f
!
Now examine the factor
ni + nf
2n2i n2f
!
ni + nf
2n2i n2f
!
and use ni > nf to argue that
1
<
n3i
<
1
n3f
(c) What do you conclude about the frequency of emitted radiation compared with the
frequencies of orbital revolution in the initial and final states? What happens as ni → ∞?
Solution
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4.4. PROBLEM 4.31 (IN THE TEXT BOOK)
13
(a) Given that the orbital frequency of an electron in an atomic orbit is:
v
2πr
and that the electron’s angular momentum is quantized,
fe =
me vr = n~
then the velocity of the electron in an atomic orbit is:
v=
n~
me r
and since the radius of an orbit is r = n2 a◦ where a◦ is the orbit with n = 1, and is given
by:
a◦ =
~2
me ke2
then the orbital frequency of the electron becomes:
fe =
=
=
=
=
v
2πr
n~
2πme r2
n~
2πme n4 a2◦
~
m2e k 2 e4 1
×
2πme
~4
n3
me k 2 e4
1
3
2π~
n3
(4.1)
(b) an electron in an orbit n has an energy En given by:
me k 2 e4
En = −
2~2
1
n2
An photon is emitted when the electron jumps from an outer orbit (ni ) to an inner orbit
(nf ), the difference in electron energy is carried by the photon as hfphoton , i.e.
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
14
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
hfphoton = Eni − Enf
fphoton
me k 2 e4
=
2~2
1
1
− 2
2
nf
ni
!
me k 2 e4
=
4π~3
1
1
− 2
2
nf
ni
!
Let us consider;
n2i − n2f
1
1
=
−
n2f
n2i
n2i n2f
(ni + nf )(ni − nf )
=
n2i n2f
2(n1 + nf )
=
(ni − nf )
2n2i n2f
The photon frequency then becomes:
fphoton
me k 2 e4
=
2π~3
ni + nf
2n2i nf 2
(ni − nf )
for adjacent orbits ni = nf + 1, so
1
1
1
=
< 3
3
3
ni
(nf + 1)
nf
(4.2)
and
ni + nf
nf + 1 + nf
2nf + 1
=
=
2 2
2
2
2ni nf
2(nf + 1) nf
2(nf + 1)2 n2f
then we need to prove that:
2nf + 1
1
<
2
2
2(nf + 1) nf
n3f
2nf + 1
1
<
2
2(nf + 1)
nf
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4.4. PROBLEM 4.31 (IN THE TEXT BOOK)
15
or
2n2f + nf < 2n2f + 4nf + 2
nf < 4nf + 2
(4.3)
Using Equations (4.2) and (4.3) we can see that:
ni + nf
1
1
<
< 3
3
2 2
ni
2ni nf
nf
(4.4)
(c) Comparing Equations (4.1) and (4.4) we find, for ni − nf = 1, that:
fei < fphoton < fef
Since ni = nf + 1, as ni gets smaller, nf also gets smaller and fei and fef get closer
together and eventually we get fei = fef = fphoton
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
16
4.5
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
Problem 4.32 (In the text book)
Wavelengths of spectral lines depend to some extent on the nuclear mass. This occurs
because the nucleus is not an infinitely heavy stationary mass and both the electron and
nucleus actually revolve around their common center of mass. It can be shown that a system
of this type is entirely equivalent to a single particle of reduced mass µ that revolves around
the position of the heavier particle at a distance equal to the electron-nucleus separation.
See Figure (4.2). Here, µ = me M/(me + M ), where me is the electron mass and M is the
nuclear mass. To take the moving nucleus into account in the Bohr theory we replace me
with µ. Thus Equation 4.30 becomes
−µke2 1
En =
2me a◦ n2
and Equation 4.33 becomes
µke2
1
=
λ
2me a◦ hc
1
1
− 2
2
nf
ni
!
=
µ
me
R
1
1
− 2
2
nf
ni
!
Determine the corrected values of wavelength for the first Balmer line (n = 3 to n = 2
transition) taking nuclear motion into account for
(a) hydrogen, 1 H,
(b) deuterium, 2 H, and
(c) tritium, 3 H. (Deuterium, was actually discovered in 1932 by Harold Urey, who measured
the small wavelength difference between 1 H and 2 H.)
Figure 4.3: (a) Both the electron and the nucleus actually revolve around the center of mass. (b)
To calculate the effect of nuclear motion, the nucleus can he considered to be at rest and m is
replaced by the reduced mass µ.
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4.5. PROBLEM 4.32 (IN THE TEXT BOOK)
17
Solution
The wavelength can be written as:
!
µ
1
1
R
− 2
2
me
nf
ni
!
1
1
R
= µ
− 2
me
n2f
ni
1
=
λ
= 1.673053 × 1036 µ
5.977098 × 10−37
λ =
µ
(a) For hydrogen atom the reduced mass is:
me M1H
me + M1H
9.109390 × 10−31 × 1.673533 × 10−27
=
9.109390 × 10−31 + 1.673533 × 10−27
= 9.104434 × 10−31 kg
µ1H =
and the wavelength is:
5.977098 × 10−37
µ1H
5.977098 × 10−37
=
9.104434 × 10−31
= 6.565041 × 10−7 m
= 656.5041 nm
λ1H =
(b) Similarly for deuterium 2 H, using M2H = 3.344495 × 10−27 kg, then reduced mass is:
me M2H
me + M2H
9.109390 × 10−31 × 3.344495 × 10−27
=
9.109390 × 10−31 + 3.344495 × 10−27
= 9.106912 × 10−31 kg
µ2H =
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
18
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
and the wave length is
5.977098 × 10−37
µ2H
5.977098 × 10−37
=
9.106912 × 10−31
= 6.563254 × 10−7 m
= 656.3254 nm
λ2H =
(c) Similarly for tritium 3 H, using M3H = 5.008234 × 10−27 kg, then reduced mass is:
me M3H
me + M3H
9.109390 × 10−31 × 5.008234 × 10−27
=
9.109390 × 10−31 + 5.008234 × 10−27
= 9.107733 × 10−31 kg
µ3H =
and the wave length is
5.977098 × 10−37
µ3H
5.977098 × 10−37
=
9.107733 × 10−31
= 6.562663 × 10−7 m
= 656.2663 nm
λ3H =
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4.6. PROBLEM 4.44 (IN THE TEXT BOOK)
4.6
19
Problem 4.44 (In the text book)
In a hot star, a multiply ionized atom with a single remaining electron produces a series of
spectral lines as described by the Bohr model. The series corresponds to electronic transitions
that terminate in the same final state. The longest and shortest wavelengths of the series
are 63.3 nm and 22.8 nm, respectively.
(a) What is the ion?
(b) Find the wavelengths of the next three spectral lines nearest to the line of longest wavelength.
Solution
(a) For a multiply ionized atoms with just one electron left, the energy of the electron is
given by:
ke2
En = −
2a◦
Z2
n2
where Z is the charge of the nucleus of the atom and R is Rydberg constant.
The wavelength of an emitted radiation from such atom is then:
1
= Z 2R
λ
1
1
− 2
2
nf
ni
!
(4.5)
The shortest wavelength λs is emitted when the electron energy change is maximum, i.e.
when ni is very large or 1/ni 1/nf , so
1
Z 2R
= 2
λs
nf
(4.6)
The longest wave λ` , is emitted when the electron energy change is minimum, i.e. when
ni = nf + 1, so
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
20
CHAPTER 4. PARTICLE NATURE OF MATTER. SOLUTIONS OF SELECTED
PROBLEMS
!
1
1
−
n2f
(nf + 1)2
"
2 #
Z 2R
nf
=
1−
n2f
nf + 1
1
= Z 2R
λ`
(4.7)
Dividing Equation (4.6) by Equation (4.7) we get:
λs
=1−
λ`
nf
nf + 1
2
and
nf
=
nf + 1
r
1−
r
1−
=
nf
λs
λ`
22.8
63.3
= 0.800
= 4
Substituting for nf = 4 into Equation (4.6) we solve for Z.
s
Z =
n2f
Rλs
r
=
1.097 ×
= 7.998
= 8
107
16
× 22.8 × 10−9
The ion has 8 charges on its nucleus it must an ion of O and since it lost all its electrons
except one then it is O7+ .
(b) The line with the longest wavelength comes from a transition from n = 5 orbit to n = 4
orbit. The nearest three lines will come form transition from n = 6, 7, 8 to n = 4,
substituting with these values in Equation (4.5) we get:
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein
4.6. PROBLEM 4.44 (IN THE TEXT BOOK)
λ = (Z 2 R)−1
21
1
1
− 2
2
nf
ni
!−1
!−1
1
1
− 2
2
nf
ni
−1
1
1
−9
−
1.424 × 10
42 62
41.10 nm
−1
1
1
−9
−
1.424 × 10
42 72
33.83 nm
−1
1
1
−9
1.424 × 10
−
42 82
30.38 nm
= 1.424 × 10−9
λ1 =
=
λ2 =
=
λ3 =
=
Physics 205:Modern Physics I, Chapter 4
Fall 2004
Ahmed H. Hussein