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Chapter 38. Modern Physics and the Atom Physics, 6 Edition
Chapter 38. Modern Physics and the Atom
Relativity
38-1. A spaceship travels past an observer at a speed of 0.85c. A person aboard the space craft
observes that it requires 6.0 s for him to walk the length of his cabin. What time would the
observer record for the same event? [ Proper time to = 6 s, relative time t =? ]
v c
c c
α = = = (^2 )
6.0 s ; 1 1- (0.85)
t o t α
t = 11.4 s
38-2. A rocket A moves past a Lab B at a speed of 0.9c. A technician in the lab records 3.50 s
for the time of an event which occurs on the rocket. What is the time as reckoned by a
person aboard the rocket? [ Proper time t 0 = ?, relative time t = 3.50 s ]
2 0 2
v c t o t t t c c
α α α
2 t 0 (^) = t 1 − (0.90) = (3.5 s) 1 − 0.81; to = 1.52 s
38-3. A blinking light on a spacecraft moves past an observer at 0.75c. The observer records that
the light blinks at a frequency of 2.0 Hz. What is the actual frequency of the blinking
light? ( It’s important to distinguish relative time from relative frequency.)
Relative frequency is 2.0 blinks per second, which is relative time of 0.5 s/blink
Relative time t = 0.50 s ; proper time to = ?;
v
c c
α = = =
2 2 ;^01 (0.75)^ (0.50 s) 1^ 0.563; 1
t o t t t α
to = 0.331 s/blink
0 0
0.331 s
f t
= = (^) fo = 3.02 Hz
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
38-4. A particle on a table has a diameter of 2 mm when at rest. What must be the speed of an
observer who measures the diameter as 1.69 mm? ( Proper length Lo = 2 mm. )
2 2 2 2 2 2 (^0 2 2 ) 0 0
L (1.69 mm) 1 ; 1 ; 1 1 L (2.00 mm)
L
L L
L
= − α = − α α = − = −
2 0.286; 0.286 0.535;
v
c
α = α= = = (^) v = 0.535 c
38-5. A blue meter stick is aboard ship A and a red meterstick is aboard ship B. If ship A moves
past B at 0.85c, what will be the length of each meterstick as reckoned by a person aboard
ship A? (W e must be careful to distinguish proper length from relative length .)
Observer A sees blue stick as proper length Lo and red stick as relative length L.
LB = 1.00 m; 2 L (^) r= (1.00 m) 1 − (0.85) ; Lr = 52.7 cm
38-6. Three meter sticks travel past an observer at speeds of 0.1c, 0.6c, and 0.9c. What lengths
would be recorded by the observer? ( Proper lengths Lo are each 1.00 m)
2 r
= 0.1; L (1.00 m) 1 (0.1 ) ;
c c c
α = = − (^) Lr = 99.5 cm;
2 r
= 0.6; L (1.00 m) 1 (0.6) ;
c
c
α = = − Lr = 80.0 cm
2 r
= 0.9; L (1.00 m) 1 (0.9) ;
c
c
α = = − (^) Lr = 43.6 cm
38-7. What mass is required to run about 1 million 100-W light bulbs for 1 year?
Eo = moc
2 = (1 x 10
6 )(100 W)(86,400 s/d)(356 d/yr) = 3.154 x 10
15 J/yr
15
2 8 2
3.154 x 10 J ; (3 x 10 m/s)
o o
E
m c
= = (^) mo = 35.0 g
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
38-11. The energy E of a photon in joules is found from the product hf. Often we are given the
wavelength of light and need to find its energy in electron volts. Show that
E
λ
such that if λ is in nanometers, E will be the energy in electron volts.
-34 8 - ; (6.63 x 10 J/Hz)(3 x 10 m/s) = 1.99 x 10 J m
hc E E λ hc λ
1 ev 1 nm (1.99 x 10 J m) 1.6 x 10 J 1 x 10 m
E λ
; Ελ = 1240 eV⋅ nm
When λ is in nm , E will be in J:
E
λ
38-12. Use the equation derived in Problem 38-13 to verify that light of wavelength 490 nm has an
energy of 2.53 eV. Also show that a photon whose energy is 2.10 eV has a wavelength of
590 nm.
490 nm
E
λ
= = E = 2.53 J ;
E 2.10 eV
λ = = λ = 590 nm
*38-13. The threshold frequency for a certain metal is 2.5 x 10
14 Hz. What is the work function?
If light of wavelength 400 nm shines on this surface, what is the kinetic energy of ejected
photoelectrons?
W = hfo = (6.63 x 10
- J/Hz)(2.5 x 10 14 Hz); W = 1.66 x 10 - J
19
1 ev 1.66 x 10 J ; 1.6 x 10 J
W
− ^
W = 1.04 eV
-34 8
(6.63 x 10 J/Hz)(3 x 10 m/s) 1.66 x 10 J 400 x 10 m
k
hc E W λ
= − = − ;^ Ek =^ 3.31 x 10
19
1 ev 3.31 x 10 J ; 1.6 x 10 J
E
− ^
Ek = 2.07 eV
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
*38-14. When light of frequency 1.6 x 10
15 Hz strikes a material surface, electrons just begin to
leave the surface. What is the maximum kinetic energy of photoelectrons emitted from this
surface when illuminated with light of frequency 2.0 x 10 15 Hz?
-34 15 15 Ek = hf − hf 0 (^) = h f ( − f 0 ); Ek = (6.63 x 10 J/Hz)(2.0 x 10 Hz −1.6 x 10 Hz);
19
1 eV 2.65 x 10 eV ; 1.6 x 10 J
E k
− ^
Ek = 1.66 eV
*38-15. The work function of nickel surface is 5.01 eV. If a nickel surface is illuminated by light of
wavelength 200 nm, what is the kinetic energy of the ejected electrons?
-34 8
hc (6.63 x 10 J/Hz)(3 x 10 m/s) ; 9.945 x 10 J 6.22 eV (200 x 10 m)
k
hc E W λ λ
k^ 6.22 eV^ 5.01 eV;
hc E W λ
= − = − (^) Ek = 1.21 eV
*38-16. The stopping potential is a reverse voltage that just stops the electrons from being emitted in
a photoelectric application. The stopping potential is therefore equal to the kinetic energy of
ejected photoelectrons. Find the stopping potential for Problem 38-15.
The kinetic energy of the emitted electrons in Prob. 38-15 is 3.31 x 10
3.31 x 10 J (1 ) ; ; (1 ) 1.6 x 10 C
k k s s s
E
E qV e V V e
= = = = (^) Vs = 2.07 V
Waves and Particles
38-17. What is the de Broglie wavelength of a proton (m = 1.67 x 10
- kg) when it is moving with
a speed of 2 x 10
7 m/s?
-27 7
(6.63 x 10 J/Hz) ; (1.67 x 10 kg)(2 x 10 m/s)
h
mv
λ = = λ = 1.99 x 10
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
*38-22. The charge on a proton is +1.6 x 10
- C and its rest mass is 1.67 x 10
27 kg. What is the de
Broglie wavelength of a proton if it is accelerated from rest through a potential difference of
500 V?
-27 - p = 2 mEk = 2 m qV ( ); p = 2(1.67 x 10 kg)(1.6 x 10 C )(500 V)
p = 5.17 x 10 kg m/s;⋅
6.63 x 10 J/Hz ; 5.17 x 10 kg m/s
h
p
λ = = ⋅
λ = 1.28 pm
Atomic Spectra and Energy Levels
38-23. Determine the wavelength of the first three spectral lines of atomic hydrogen in the Balmer
series_. (For the Balmer series, n = 2. The first three lines come for ni =3, 4, and 5.)_
2 2
(^2) i
R
λ n
; ( ni = 3, 4, and 5)
(a)
7 - 2
(1.097 x 10 m ) ; λ 4 3
λ = 656.3 nm ( ni = 3)
(b)
7 - 2
(1.097 x 10 m ) ; λ 4 4
λ = 486.2 nm ( ni = 4)
(c)
7 - 2
(1.097 x 10 m ) ; λ 4 5
λ = 434.1 nm ( ni = 5)
38-24. Find the wavelengths of the first three lines of atomic hydrogen in the Paschen series.
(a)
7 - 2
(1.097 x 10 m ) ; λ 9 4
λ = 1887 nm ( ni = 4)
(b)
7 - 2
(1.097 x 10 m ) ; λ 9 5
λ = 1290 nm ( ni = 5)
(c)
7 - 2
(1.097 x 10 m ) ; λ 9 6
λ = 1101 nm ( ni = 6)
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
38-25. Determine the radius of the n = 4 Bohr level of the classical Bohr hydrogen atom.
2 2 2 -12 2 2 -34 2 0 2 -34 -
(4) (8.85 x 10 C /N m )(6.63 x 10 J/Hz) ; (9.1 x 10 J/Hz)(1.6 x 10 )
n h r me C
ε
π π
= = r = 847 pm
38-26. What is the classical radius of the first Bohr orbit in the hydrogen atom?
2 2 2 -12 2 2 -34 2 0 2 -34 -
(1) (8.85 x 10 C /N m )(6.63 x 10 J/Hz) ; (9.1 x 10 J/Hz)(1.6 x 10 C)
n h r me
ε
π π
= = (^) r = 53.2 pm
38-27. Determine the wavelength of the photon emitted from a hydrogen atom when the electron
jumps from the n = 3 Bohr level to ground level.
7 - 2 2 2 2
(1.097 x 10 m ) ; (^2) i 1 3
R
λ n
^
λ = 102.6 nm
*38-28. What is the maximum wavelength of an incident photon if it can ionize a hydrogen atom
originally in its second excited state (n = 3)?
Ionization Energy:
3 2 2 3
13.6 eV 13.6 eV ; 1.511 eV 2.42 x 10 J 3
E E
n
-34 8
3
(6.63 x 10 J/Hz)(3 x 10 m/s) ; 2.42 x 10 J
hc
E
λ = = (^) λ = 823 nm
*38-29. What are the shortest and longest possible wavelengths in the Balmer series?
( Since E=hc/ λ , the highest energy (greatest n level) is
shortest wavelength and vice versa.)
Shortest λ , n i = ∞:
7 - 2
(1.097 x 10 m ) ; λ 2
λ = 365 nm
Longest λ , n i = 3:
7 - 2 2
(1.097 x 10 m ) ; λ 2 3
λ = 656 nm
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
38-33. In the hydrogen atom an electron falls from the n = 5 level to the n = 2 level and emits a
photon in the Balmer series. What is the wavelength and energy of the emitted light?
7 - 2 2
(1.097 x 10 m ) ; λ 2 5
λ = 434 nm
-34 8
(6.63 x 10 J/Hz)(3 x 10 m/s)
434 x 10 m
hc E λ
= = ; E = 4.58 x 10
38-34. Calculate the frequency and the wavelength of the Hβ line of the Balmer series. The
transition is from the n = 4 level of the Bohr atom.
7 - 2 2
(1.097 x 10 m ) ; λ 2 4
λ = 486.2 nm ( ni = 4)
8
(3 x 10 m/s) ; 486 x 10 m
c f λ
= = f = 6.17 x 10
14 Hz
38-35. A spaceship A travels past another ship B with a relative velocity of 0.2c. Observer B
determines that it takes a person on ship A exactly 3.96 s to perform a task. What time will
be measured for the same event by observer A? (Event happens aboard A)
Observer A records proper time to; Observer B records relative time t = 3.96 s.
2 0 2
; 1 (0.20) (3.96 s) 1 0.04; 1
t o t t t α
to = 3.88 s
*38-36. The rest mass of an electron is 9.1 x 10
- kg. What is the relativistic mass of an electron
traveling at a speed of 2 x 10 8 m/s? What is the total energy of the electron? What is its
relativistic kinetic energy?
8
(^8 2 )
2 x 10 m/s 1 kg 0.667; ; 3 x 10 m/s (^1 1) (0.667)
m o α m α
m = 16.4 x 10
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
*38-36. (Cont.) E = mc
2 = (16.4 x 10
8 m/s)
2 ; E = 1.47 x 10
Ek = (m – mo)c
2 ; Ek = (16.4 x 10
- kg – 9.1 x 10 - kg) (3 x 10
8 m/s)
2 ;
Ek = 6.57 x 10
*38-37. What is the de Broglie wavelength of an electron whose kinetic energy is 50 MeV?
1.6 x 10 J - 50 MeV 8.00 x 10 J; 1 MeV
E k
2 k^ ½^ ;
p E mv v m
2 -31 - k^ ½^ ;^2 k 2(9.1 x 10^ kg)(8 x 10^ J);
p E m p mE m
6.63 x 10 J/Hz 3.82 x 10 kg m/s; ; 3.82 x 10 kg m/s
h p p
= ⋅ λ= = ⋅
λ = 0.174 pm
*38-38. The rest mass of a proton is 1.67 x 10
- kg. What is the total energy of a proton that has
been accelerated to a velocity of 2.5 x 10
8 m/s? What is its relativistic kinetic energy?
8 8
8 8
2 x 10 m/s 2 x 10 m/s 0.667; 0.833; 3 x 10 m/s 3 x 10 m/s
α= = α= =
2 -27 8 2
2
(1.67 x 10 kg)(3 x 10 m/s) E ; 1 (0.833) 0.
m c o = = −
E = 2.72 x 10
Ek = E - moc
2 ; Ek = 2.72 x 10
- J – (1.67 x 10 - kg)(3 x 10
8 m/s)
2 ;
Ek = 1.22 x 10
*38-39. Compute the mass and the speed of protons having a relativistic kinetic energy of 235
MeV. The rest mass of a proton is 1.67 x 10
-11 2 2 2 0
1.6 x 10 J 235 MeV 3.76 x 10 J; ; 1 MeV
k k k o
E
E E mc m c m m c
8 2
3.76 x 10 J 1.67 x 10 kg; (3 x 10 m/s)
m = + m = 2.09 x 10
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
*38-41. A particle of mass m is traveling at 0.9c. By what factor is its relativistic kinetic energy
greater than its Newtonian kinetic energy? ( First ignore effects of relativity)
Ek = ½mov
2 = ½ mo (0.9c)
2 ; Ek = 0.405 moc
2 Now with relativity:
(^2 ) 2 0 2 2 2 2 0 0 0 0
k k
c m c E m c m c m c E m c c
α
2 0 2 0 0
E rel m c ratio E m c
= = (^) ratio = 3.
*38-42. What is the momentum of a 40-eV photon? What is the wavelength of an electron with the
same momentum as this photon?
8
(40 eV)(1.6 x 10 J/eV) ; ; (3 x 10 m/s)
E
E pc p c
= = = (^) p = 2.13 x 10
(6.63 x 10 J/Hz) ; 2.13 x 10 kg m/s
h
p
λ = = ⋅
λ = 31.1 nm
*38-43. When monochromatic light of wavelength 410 nm strikes a cathode, photoelectrons are
emitted with a velocity of 4.0 x 10
5 m/s. What is the work function for the surface and
what is the threshold frequency?
2 -31 5 2 - Ek = ½ mv = ½(9.1 x 10 kg)(4 x 10 m/s) ; Ek =7.28 x 10 J
-34 8 -20 -
(6.63 x 10 J/Hz)(3 x 10 m/s) 7.28 x 10 J = 4.12 x 10 J; 410 x 10 m
k
hc W E λ
19
1 eV 4.12 x 10 ; 1.6 x 10 J
W
− ^
W = 2.58 eV
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
(^0 0) -
4.12 x 10 J ; ; (6.63 x 10 J/Hz)
W
W hf f h
= = = (^) fo = 6.23 x 10 14 Hz
*38-44. What is the velocity of a neutron (m = 1.675 x 10
- kg) that has a de Broglie wavelength
of 0.1 nm. What is its kinetic energy in electronvolts?
-27 -
(6.63 x 10 J/Hz) ; (1.675 x 10 kg)(0.1 x 10 m)
h h v mv m
λ λ
= = = (^) ; v = 3960 m/s
Ek = ½(1.675 x 10
2 ; Ek = 1.31 x 10
*38-45. What is the velocity of a particle whose relativistic kinetic energy is twice its rest mass
energy? [ Note: The conditions are that Ek(rel.) = 2moc
2 ]
(m - mo)c
2 =2moc
2 ; mc
2
- moc
2 = 2moc
2 ; mc
2 = 3moc
2 ; m = 3mo
0 2 2 2 2 0
m m α α α α
v v c c
α = = = (^) v = 2.83 x 10^8 m/s
*38-46. Compute the relativistic mass and the speed of electrons having a relativistic kinetic
energy of 1.2 MeV.
1.6 x 10 J 1.20 MeV 1.92 x 10 J; ; 1 MeV
k k k o
E
E E mc m c m m c
8 2
1.92 x 10 J 9.1 x 10 kg; (3 x 10 m/s)
m = + m =^ 3.04 x 10
2 2 0 2 0 2 0 2 0 2
m m m m m m m m
α α α α
− ^ ^ ^
-31^2 2
9.1 x 10 kg 1 ; 0.910 0.954; 3.04 x 10 kg
α α
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
(a)
2 2 Newtonian Work = ½(1 kg)(0.2c) − ½(1 kg)(0.1c) Newtonian Work = 1.35 x 10 15 J
*38-48. (Cont.) Relativistic work:
2 0 2 f 0
f f o
v (^) c v c
c c c c
α = = α = α = = α =
2 2 2 2 0 0 0 0 2 2 2 0 0
1 1 1 0.04^1 0.
k f
m c m c m c m c E Work m c α α
− − −^ −
Relativistic Work = 1.40 x 10
15 J; Newtonian Work = 1.35 x 10
15 J
(b)
2 2 Newtonian Work = ½(1 kg)(0.8c) − ½(1 kg)(0.7c) Newtonian Work = 6.75 x 10 15 J
Relativistic work:
2 0 2 f 0
f f o
v (^) c v c
c c c c
α = = α = α = = α =
2 2 2 2 0 0 0 0 2 2 2 0 0
1 1 1 0.64^1 0.
k f
m c m c m c m c E Work m c α α
− − −^ −
Relativistic Work = 24.0 x 10
15 J; Newtonian Work = 6.75 x 10
15 J
*38-49. An electron in the hydrogen atom drops from the n = 5 level to the n = 1 level. What are
the frequency, wavelength, and energy of the emitted photon. In which series does this
photon occur? How much energy must be absorbed by the atom in order to kick the
electron back up to the fifth level?
7 - 2 2
(1.097 x 10 m ) ; λ 1 5
λ = 95.0 nm
8
(3 x 10 m/s) ; 95.0 x 10 m
c f λ
= = f = 3.16 x 10
15 Hz
-34 15 E = hf = (6.63 x 10 J/Hz)(3.16 x 10 Hz); E = 2.09 x 10
The series that has n = 1 as its final level is: Lyman Series:
The energy absorbed must be the same: Eabs = 2.09 x 10
Chapter 38. Modern Physics and the Atom Physics, 6 Edition
*38-50. In a photoelectric experiment shown as Fig. 38-15, a source of emf is connected in series
with a galvanometer G. Light falling on the metal cathode produces photoelectrons. The
source of emf is biased against the flow of electrons, retarding their motion. The potential
difference V 0 just sufficient to stop the most energetic photoelectrons is called the stopping
potential. Assume that a surface is illuminated with light of wavelength 450 nm. causing
electrons to be ejected from the surface at a maximum speed of 6 x 10
5 m/s. What is the
work function for the surface and what is the stopping potential?
(a) Ek = ½mv
2 = ½(9.1 x 10
5 m/s)
2 ; Ek, = 1.64 x 10
-34 8
(6.63 x 10 J/Hz)(3 x 10 m/s) 1.64 x 10 J; 450 x 10 m
k
hc W E λ
= − = − W = 2.78 x 10
(b) The stopping potential must provide energy equal to (1e)Vo = Ek:
(^0) -
1.64 x 10 J 1.02 V; 1.6 x 10 J/V
V = = Vo = 1.02 V
*38-51. In a photoelectric experiment, 400-nm light falls on a certain metal, and photoelectrons
are emitted. The potential required to stop the flow of electrons is 0.20 V. What is the
energy of the incident photons? What is the work function? What is the threshold
frequency?
Work = ∆ Ek = ½mv
2 ; ∆ Ek = qV = (1e)V So that Ek = eVo
Ek = eVo = hf – W; where Vo is stopping potential
-34 8 19
(6.63 x 10 J/Hz)(3 x 10 m/s) 4.97 x 10 J; 400 x 10 m
hc E λ
− = = = E = 3.11 eV
W = E – eV = 3.11 eV – 0.20 eV; W = 2.91 eV
