AP Physics 6: Subatomic Mass, Energy & Momentum Name __________________________
A.
Atomic Nucleus
1. 2 types of particles—nucleons
a. positively charged proton and neutral neutron
b. rest mass chart (1 u = 1.66 x 10-27 kg)
Object
kg
u
Proton (11p)
1.67262 x 10-27
1.007276
Neutron (10n)
1.67493 x 10-27
1.008665
0
Electron ( -1e)
9.1094 x 10-31
0.00054858
2. nuclide (combination of protons and neutrons)
a. number of protons = atomic number (Z)
b. number of protons + neutrons = mass number (A)
c. nuclide symbol: AZX
1. X is atomic symbol with same Z number
2. A varies = isotopes
3. binding energy (BE) and nuclear forces
a. takes energy to break a stable nuclide into its
parts  energy is added to the nuclide to
separate into protons and neutrons
1. nuclide + BE = protons + neutrons
2. change in mass, E = mc2
a. c = 3 x 108 m/s (speed of light)
b. if E = BE, then m = mBE in kg
c. mnuclide + mBE = mp + mn
b. binding energy per nucleon (BE/# nucleons)
c.
4.
nuclear processes
1. exothermic when BE/#nucleon increases
2. mproducts – mreactants = mBE (-mBE is exothermic)
3. nuclear fusion: Zsmall  Zlarge
a. thermonuclear devise—Hydrogen bomb
b. sun's energy
4. nuclear fission: Zlarge  Zsmall
a. atomic bomb (Hiroshima)
b. nuclear power plants
strong nuclear force
a. acts over a very short distance (10-18 m)
b. bind "up" and "down" quarks together
1. proton = 2 up (u+2/3) + 1 down (d-1/3) quark
2. neutron = 1 up (u+2/3) + 2 down (d-1/3) quark
c.
d.
proton
neutron
binds neutrons and protons together
neutrons stabilize nuclide
1. electric repulsion between protons is
significant when protons > 10-18 m apart
2. neutrons increase total strong nuclear force
without increasing electric repulsion
5.
B.
naturally occurring radioactivity
a. unstable nuclides undergo nuclear change
b. nuclide is too large (Z > 84)
1. alpha radiation, 42 (42He)—low penetration
2. 22688Ra  22286Rn + 42
3. BE/# nucleons increases
c. A > average atomic mass
1. beta radiation, 0-1 (0-1e)—medium penetration
a. -⅓d  ⅔u + 0-1
b. 10n  11p + 0-1c. involves "weak nuclear force" and
generates an additional particle called
neutrino ()
2. 146C  147N + 0-1
d. A < average atomic mass
1. positron radiation, 01
a. ⅔u  -⅓d + 01
b. 11p  10n + 01
2. 116C  115B + 01
e. nuclide falls from high energy to low energy state
1. gamma radiation, 00—high penetration
2. matter-antimatter collision
a. 0-1e + 01e  
b. E = 2mec2
6. artificial induced radioactivity (transmutation)
a. nuclide bombarded by subatomic particle (proton,
neutron, alpha particle, etc.)
b. nuclear reaction
1. mass and charge are conserved
2. 10n + 147N  146C + 11p
7. rate of decay and half-life
a. time it takes to reduce radioactivity by half is
constant = half life, t½
b. 1  1/2  1/4  1/8  1/16 ...
Photons and Electrons
1. photon
a. quantum unit of electromagnetic energy
b. wave property, c = f
1. wavelength,  (m)
2. frequency, f (s-1)
c. energy, E = hf
Planck's constant, h = 6.63 x 10-34 J•s
electromagnetic radiation
a. gamma > x-rays > UV > visible > IR > radio
b. violet > blue > green > yellow > orange > red
3. brightness measures intensity (not energy)
4. warm objects glow infrared (hot = visible light)
d. relativistic mass, m = E/c2 (mo = 0)
e. momentum, p = mc
f. interconnecting formulas (kg, m, s, J)
m
f
E/p
In terms of:

E = mc2
E = hf
E = pc
Energy
E = hc/
p = mc
p = hf/c
p = E/c
Momentum
p = h/
1.
2.
electron
a. electron properties
1. same charge as proton, but negative
2. mo = 9.11 x 10-31 kg (0.00054858 u)
b. electron structure in an atom: Bohr model
1.
c.
electrons occupy discrete energy levels
a. En = -B/n2 (BH = 13.6 eV)
b. n = energy level #
c. electron volt (eV) = small unit of energy
1 eV = 1.6 x 10-19 J
2. electrons change n by absorbing/emitting
photons
a. Ephoton = En-high – En-low
b. EeV = 1240 eV•nm/nm
photoelectrons
1. Ephoton > ionization energy  atom emits
electron
2. Einstein's photoelectric effect equation
a.
Kelectron = Ephoton – 
Kelectron
2.
threshold
b.
d.
fphoton
 called "work function" = ionization
energy
1. minimum energy for photoelectrons
2. threshold nm = 1240 eV•nm/eV
c. electron can be stopped by a reverse
voltage
1. called stop voltage, Vstop
2. Vstop = Kelectron (in eV)
3. (KeV)(1.6 x 10-19) = KJ
KJ = ½(9.11 x 10-31 kg)v2
De Broglie wavelength: particle = h/p = h/mov
A. Atomic Nucleus
Questions 1-21 Briefly explain your answer.
1. There are 82 protons in a lead nucleus. Why doesn't the
lead nucleus burst apart?
(A) electric repulsive forces don't exist in the nucleus
(B) gravity overpowers the electric repulsion
(C) neutrons neutralize the positively charged protons
(D) none of the above
D Strong nuclear forces > electric repulsion
C 90 years = 3 t½ (16 g  8 g  4 g  2 g)
15. How long will it take to reduce the amount to 8 g?
(A) 30 yr
(B) 60 yr
(C) 90 yr
(D) 120 yr
A 16 g  8 g = 1 t½ (30 yr)
16. What is the half life of a radioactive material that decays to
¼ the original amount in 16 years?
(A) 1 yr
(B) 2 yr
(C) 4 yr
(D) 8 yr
D 1  ½  ¼ (2 t½)  16 yr/2 = 8 yr
What weighs more, A—an electron and a proton, or B—a
hydrogen atom?
17. Which is more radioactive, substances A (t½ = 100 s) or B
(A) A
(B) B
(C) tie
(t½ = 50 s)?
It takes energy to remove an electron from neutral atom 
(A) A
(B) B
(C) tie
A
separate parts have more total mass
B Shorter t½ equals more radioactive
Questions 3-5 radiation passes through a vacuum, where one
side is positively charged and the other is negatively
18. Which type of radiation goes farther in matter before losing
charged.
++++++++++++++++++++
all of its energy?
A
(A) alpha
(B) beta
(C) gamma
Gamma radiation is pure energy  with charge or mass, it
B
C
can penetrate matter without much loss in energy
______________ C
3. Which is the path of an alpha particle?
19. Define
2.
C Negative charge attracts + alpha particle
4.
Which is the path of a beta particle?
A Positive charge attracts – beta particle
5.
Which is the path of gamma radiation?
B Gamma is uncharged  not attracted to + or – charge
Questions 6-9 Refer to the nonspontaneous nuclear decay:
238 U  234 Th + 4 He
92
90
2
6. Which weighs more, A—U-238, or B—Th-234 + He-4?
(A) A
(B) B
(C) tie
Nonspontaneous  energy (mass) is added to reactants
B
form products
7. Which has greater momentum, A—Th-234, or B—He-4?
(A) A
(B) B
(C) tie
C Equal impulse (Ft) at separation  equal momentum
8.
Which has greater velocity, A—Th-234, or B—He-4?
(A) A
(B) B
(C) tie
Same momentum, but He has less mass  greater
B
velocity
9. Which has greater kinetic energy, A—Th-234, or B—He-4?
(A) A
(B) B
(C) tie
mHevHe = mThvTh, but in kinetic energy, v is squared  He
B
has more kinetic energy
10. What is produced by the beta decay of 31H?
(A) 21H
(B) 11H
(C) 32He
(D) 42He
C
3
1H
 0-1 + 32He
B
84Po
 2 +
4
206 Pb
82
12. Complete the nuclear equation: 146C + 0-1e  ___
(A) 156C
(B) 157N
(C) 145B
(D) 147N
C
14
6C
+ 0-1e  145B
C
0n
+
16 O
8

15
7N
nuclide
Combination of protons and neutrons
Z#
Atomic number = number of protons
A#
Mass number = number of protons + neutrons
isotope
same Z value, but different A value
20. Consider a carbon-12 atom (a.m. = 12.0 u).
a. What is the mass in kg?
12.0 u x 1.66 x 10-27 kg/1 u = 1.99 x 10-26 kg
b.
What is the energy equivalence in J?
E = mc2 = (1.99 x 10-26 kg)(3 x 108 m/s)2 = 1.79 x 10-9 J
21. Complete the following chart for the isotope, U-235.
Number of Number of Number of
Z value
A value
nucleons
protons
neutrons
235
92
143
92
235
22. Consider the helium nuclide (42He = 4.001504). Determine
a. the difference in mass between nucleons and nuclide.
(1) in u
m = 2(mp) + 2(mn) – mHe
m = 2(1.007276) + 2(1.008665) – 4.001504 = 0.030378 u
(2) in kg
b.
the total binding energy in J.
E = mc2 = (5.04 x 10-29 kg)(3 x 108 m/s)2 = 4.54 x 10-12 J
c.
the binding energy per nucleon.
4.54 x 10-12 J/4 = 1.13 x 10-12 J
13. Complete the nuclear equation: 10n + 168O  ___ + 21H
(A) 178O
(B) 158O
(C) 157N
(D) 159F
1
Subatomic particle found in the nucleus
0.030378 u x 1.66 x 10-27 kg/1u = 5.04 x 10-29 kg
11. What is produced by the alpha decay of 21084Po?
(A) 21082Pb (B) 20682Pb (C) 21086Rn (D) 21486Rn
210
nucleon
+
2
1H
Questions 14-15 16 g of radioactive material has a 30-yr half-life.
14. How much is left after 90 years?
(A) 8 g
(B) 4 g
(C) 2 g
(D) 1 g
23. Complete the following chart.
Symbol
Mass Charge
4 He
4
2
alpha
2
beta
gamma
0
Penetrating ability
low
-1e
0
-1
medium

0
0
high
24. Fill in the missing symbols.
10B
+ n  +
7Li
1H
+ n  +
c.
2H
50Cr
+ n  +
25. Consider the alpha decay of U-238:

+ 234Th
238U = 238.050784 u 4He = 4.002602 u 234Th = 234.055381 u
a. Determine the change in mass per U-238 atom.
(1) in u
238U
4He
m = mHe + mTh - mU
m = 4.002602 u+234.055381 u–238.050784 u=0.007199 u
(2) in kg
0.007199 u x 1.66 x 10-27 kg/1 u = 1.20 x 10-29 kg
b.
Is the reaction spontaneous?
m > 0  reaction is not spontaneous.
c.
Determine the change in energy (in J) per atom.
E = mc2
E = (1.20 x 10-29 kg)(3 x 108 m/s)2 = 1.08 x 10-12 J
26. A sample of 100Pd (half-life 4 days) has mass 1.776 g at
3:00 P.M. on July 4; what mass of 100Pd remains at 3:00
P.M. on July 16?
12 days = 3 t½
(1/8)1.776 g = 0.222 g
27. Consider a helium-4 atom (atomic mass = 4.00 u).
a. What is the mass in kg?
4.00 u x 1.66 x 10-27 kg/1 u = 6.64 x 10-27 kg
b.
What is the energy equivalence in J?
d.
If all the released energy becomes kinetic energy,
what is K/KRa?
mRavRa = mv (222)(4) = (4)(222)
K = ½mv2 = (4)(222)2 = 222 = 55.5
KRa ½mRavRa2 (222)(4)2
4
e. What is the alpha particle's velocity if 98 % of the
energy is absorbed by the particle?
E = ½mv2
(.98)(7.81 x 10-13 J) = ½(4 x 1.67 x 10-27 kg)v2 
v = 1.51 x 107 m/s
f. What is the De Broglie wavelength of the alpha particle?
 = h/mv
 = 6.63 x 10-34 J•s/(4 x 1.67 x 10-27 kg)(1.51 x 107 m/s)
 = 6.57 x 10-15 m
32. How long does it take for a sample of 32P (t1/2 = 14 days) to
lose 7/8 of its activity?
The loss of 7/8 activity means that 1/8 remains
It takes 3 half-lives to reach 1/8 (1  1/2  1/4  1/8)
3 x 14 = 42 days
B. Photons and Electrons
33. Check which color has the greatest value for the following.
red
yellow
violet
can't tell
frequency
wavelength
energy
relativistic mass
momentum
intensity
34. Complete the chart for an X-ray photon ( = 1.54 x 10-10 m).
E = mc2
E = (6.64 x 10-27 kg)(3 x 108 m/s)2 = 5.98 x 10-10 J
28. Complete the following chart for the isotope, Rn-226.
Number of Number of Number of
Z value
A value
nucleons
protons
neutrons
226
86
140
86
226
29. Determine the binding energy (in joules) of an average
nucleon in 3H (3.015500 u) by completing the following
sequence of calculations
m = 1mp + 2mn – mH-3
m
m = (1.007276) + 2(1.008665) – (3.015500)
(u)
m = 0.009106 u
m
(kg)
BE
f
c = f
3 x 108 m/s = f(1.54 x 10-10 m)f = 1.95 x 1018 s-1
E
E = hf
E = (6.63 x 10-34 J•s)(1.95 x 1018 s-1) = 1.29 x 10-15 J
m
E = mc2
1.29 x 10-15 J = m(3 x 108 m/s)2 m = 1.44 x 10-32 kg
p
p = h/
p = (6.63 x 10-34 J•s)/(1.54 x 10-10 m) = 4.31 x 10-24 kg•m/s
35. A proton and antiproton (m = 1.67 x 10-27 kg) collide and
convert all mass into photon energy. Determine the photon's
m = 0.009106 u x 1.66 x 10-27 kg/1 u
m = 1.51 x 10-29 kg
E
E = mc2= 2(1.67 x 10-27 kg)(3.0 x 108 m/s)2
E = 3.0 x 10-10 J
BE = mc2 = (1.51 x 10-29 kg)(3 x 108 m/s)2
BE = 1.36 x 10-12 J
f
E = hf  f = E/h
f = 3.0 x 10-10 J/6.63 x 10-34 J•s = 4.5 x 1023 s-1

c = f   = c/f
 = 3 x 108 m/s/4.5 x 1023 s-1 = 6.7 x 10-16 m
p
p = h/ = (6.63 x 10-34 J•s)/(6.7 x 10-16 m)
p = 1.0 x 10-18 kg•m/s
BE/A BE/A = 1.36 x 10-12 J/3 = 4.53 x 10-13 J/nucleon
30. Fill in the missing symbols.
27Al
+  1n + 30P
9Be
+  n + 12C
31. Stationary Radium-226 (225.97709 u) undergoes an alpha
decay (4.001504 u) to produce radon-222 (221.97036).
a. Write a nuclear equation for this radioactive decay.
226
b.
88Ra
Calculate the energy is joules released per decay.
E = mc2
E = (8.678 x 10-30 kg)(3 x 108 m/s)2 = 7.81 x 10-13 J
51Cr
 22286Rn + 42He
Calculate the change in mass in kg for this reaction.
221.97036 + 4.001504 – 225.97709 = -0.005226 u
-0.005226 u x 1.66 x 10-27 kg/1 u = -8.675 x 10-30 kg
Questions 36-47 Briefly explain your answer.
36. What is the ionization energy of ground state hydrogen?
(A) 0 eV
(B) 13.6 eV (C) 41.2 eV (D) 54.4 eV
B E – E1 = 0 – (-13.6 eV/12) = 13.6 eV
37. Which transition results in the greatest gain in energy?
(A) 2  5 (B) 5  2 (C) 3  10 (D) 1  2
The energy gap between 1 and 2 is ¾ of the total energy
D
between ground state and ionization
38. The Balmer series includes two blue, a blue-green and red
spectral line. Which transition generates the red?
(A) 3  2 (B) 4  2 (C) 5  2 (D) 6  2
Red would have the least energy  the electron would
A
come from the energy level closest to 2
39. A hydrogen electron is excited to n = 4. How many
different transitions are possible to reach ground state?
(A) 1
(B) 2
(C) 3
(D) 6
49. Complete the chart for a hydrogen transition from n = 2  3.
E2
E2 = -B/n2 = -13.6 eV/22 = -3.4 eV
E3
E3 = -B/n2 = -13.6 eV/32 = -1.5 eV
E
E = -1.5 eV – (-3.4 eV) = 1.9 eV
D 4  3, 4  2, 4  1, 3  2, 3  1, 2  1
B Greater wavelength equals less energy (E = hc/)  B
41. Which metal has the greater threshold frequency?
(A) A
(B) B
(C) tie
A Greater frequency equals more energy (E = hf) A
42. Green light does NOT generate photoelectrons in metal A.
Which color might be able to generate photoelectrons?
(A) red
(B) orange (C) yellow (D) blue
D red < orange < yellow < green < blue < violet blue
Questions 43-44 A metal surface emits photoelectrons when
exposed to 400-nm light.
43. What happens if the metal is exposed to 300-nm light?
(A) more electrons are emitted
(B) fewer electrons are emitted
(C) more energetic electrons are emitted
(D) no electrons are emitted
300-nm light has more energy than 400-nm light; the extra
C
energy becomes electron kinetic energy
44. What will happen if the metal is exposed to brighter 400-nm
light?
(A) more electrons are emitted
(B) fewer electrons are emitted
(C) more energetic electrons are emitted
(D) no electrons are emitted
Brighter, more intense light, has more photons, but not
A
more energetic photons
45. The speed of proton A is faster than proton B. Which one
has the longer De Broglie wavelength?
(A) A
(B) B
(C) tie
B  = h/mv    1/v  slower = longer wavelength
46. A proton and electron have the same speed, which has the
longer De Broglie wavelength?
(A) proton
(B) electron
(C) tie
B  = h/mv    1/m  lighter = longer wavelength
47. A proton and electron have the same momentum, which
has the longer De Broglie wavelength?
(A) proton
(B) electron
(C) tie
C  = h/p    1/p same p = same wavelength
48. Red light has a wavelength of 750 nm. Determine the
c = f
f
3 x 108 m/s = f(750 x 10-9 m)f = 4.0 x 1014 s-1
photon
Ephoton = (1240 eV•nm)/nm
1.9 eV = (1240 eV•nm)/nm  nm = 653 nm
50. The hypothetical atom has four energy states as shown.
a. Calculate the change in energy for all transitions from
the forth energy level to ground state.
Transition
E
n=4
-4
n = 4  n = 3 -16 - -4 = -12 eV
n=3
-16
n = 4  n = 2 -36 - -4 = -32 eV
Energy (eV)
Question 40-42 Metal A has a greater work function 
compared to metal B.
40. Which metal has the greater threshold wavelength?
(A) A
(B) B
(C) tie
-36
n=2
n = 4  n = 1 -144 - -4 = -140 eV
n = 3  n = 2 -36 - -16 = -20 eV
n = 3  n = 1 -144 - -16 = -128 eV
-144
b.
n=1
n = 2  n = 1 -144 - -36 = -108 eV
Ground state atoms are irradiated with photons having
energies between 120 eV and 130 eV. What energies
can be emitted by atoms of this gas?
20 eV, 128 eV, 108 eV
51. Consider an electron transition from the second energy
level to ground state on a hydrogen atom.
a. What are E1 and E2?
E1 = -13.6 eV/12 = -13.6 eV
E2 = -13.6 eV/22 = -3.4 eV
b.
What is E of the emission line?
E = En-high – En-low
E = -3.4 – (-13.6) = 10.2 eV
c.
What is the wavelength of the photon?
Ephoton = 1240/nm
nm = 1240/10.2 = 122 nm
d.
How much energy is needed to ionize an electron in
the n = 2 state?
E = E – En
E = 0 – (-B)/n2 = 13.6 eV/22 = 3.4 eV
52. What is the threshold wavelength that will emit electrons
from a metal whose work function is 3.10 eV?
nm = 1240 eV•nm /eV = 1240 eV•nm /3.10 eV = 400 nm
53. A metal surface ( = 2.40 eV) is exposed to 240 nm light.
a. What is the energy of the photons?
E = 1240/nm = 1240 eV•nm/240 nm = 5.17 eV
E
E = hf
E = (6.63 x 10-34 J•s)(4.0 x 1014 s-1) = 2.7 x 10-19 J
b.
What is the maximum kinetic energy (in eV) of
electrons ejected from a surface?
m
E = mc2
2.7 x 10-19 J = m(3 x 108 m/s)2 m = 2.9 x 10-36 kg
K = Ephoton -  = 5.17 eV – 2.40 eV = 2.77 eV
p
p = h/
p = (6.63 x 10-34 J•s)/(750 x 10-9 m) = 8.8 x 10-28 kg•m/s
54. When 230-nm light falls on a metal, the photoelectrons are
stopped with a voltage of 1.64 V.
a. What is the kinetic energy of the photoelectrons?
Vstop = Kelectron = 1.64 eV
b.
What is the energy of the photons?
Ephoton = 1240/nm = 1240 eV•nm/230 nm = 5.39 eV
c.
What is the work function of the metal?
 = Ephoton – K = 5.39 – 1.64 = 3.75 eV
55. A sodium surface ( = 2.28 eV) is illuminated by 500 nm
light.
a. What is the energy per photon?
E = 1240 eV•nm/500 nm = 2.48 eV
b.
What is the maximum kinetic energy of the
photoelectrons?
K = Ephoton - = 2.48 eV – 2.28 eV = 0.20 eV
c.
What is the threshold wavelength for photoemission
from the surface of sodium?
 = 1240 eV•nm/nm = 1240 eV•nm/2.28 nm = 544 nm
d
What is the kinetic energy of the photoelectrons when
600 nm light shines on the sodium surface?
0 eV, the energy of the photon is less than 
56. Determine the following with a wavelength of 500 nm.
photon
electron

(m)
p
(kg•m/s)
m
(Kg)
500 nm x 10-9 m/1 nm = 500 x 10-9 m
p = h/
p = 6.63E-34/500E-9
p = 1.3E-27kg•m/s
m = p/c
m = 1.3E-27/3E8
m = 4.4E-36 kg
p = h/
p = 6.63E-34/500E-9
p = 1.3E-27 kg•m/s
9.1E-31 kg
v = p/m
v = 1.3E-27/9.1E-31
v = 1.4E3 m/s
f = c/
f = v/
f
8/500E-9
f
=
3E
f = 1.4E3/500E-9
(s-1)
14
f = 6E
f = 2.8E9
E = hf
E = ½mv2
E
E = (6.63E-34)(6E14)
E = ½(9.1E-31)(1.4E3)2
(J)
-19
E = 4E J
E = 8.9E-25 J
4E-19 J x 1 eV
8.9E-25 J x 1 eV
E
1.6E-19 J
1.6E-19 J
(eV)
E = 2.5 eV
E = 5.6E-6 eV
Practice Multiple Choice (No calculator)
Briefly explain your answer.
1. The nuclear reaction X  Y + Z occurs spontaneously. If
Mx, My, and Mz are the masses of the three particles, which
of the following relationships is true?
(A) MX < MY – MZ
(B) MX < MY + MZ
(C) MX > MY + MZ
(D) MX – MY < MZ
Spontaneous process loses energy (mass) in order to
C
reach a lower energy state  MX > MY + MZ
2. A negative beta particle is emitted during the radioactive
decay of 21482Pb. Which is the resulting nucleus?
v
(m/s)
3E8 m/s
(A) 21080Hg (B) 21481Tl (C) 21383Bi (D) 21483Bi
0
x
82Pb  -1 + yZ
D
214 = 0 + x  x = 214, 82 = -1 + y  y = 83  21483Bi
3. The deuteron (21H) mass md is related to the neutron mass
mn and the proton mass mp by which of the following?
(A) md = mn + mp
(B) md = mn + mp + mBE
(C) md = 2(mp)
(D) md = mn + mp – mBE
Taking a deuteron apart takes energy  mBE, is added to
D
the reactant side. md + mBE = mp + mn
4. At noon the decay rate is 4,000 counts/minute. At 12:30
P.M. the decay rate is 2,000 counts/minute. The predicted
decay rate in counts/minute at 1:30 P.M. is
(A) 0
(B) 500
(C) 667
(D) 1,000
The counts/minute decrease by half in 30 min (t½)  at
B
1:00 the count is 1,000 and at 1:30 the count is 500
5. In the photoelectric effect, the maximum speed of electrons
emitted by a metal surface when it is illuminated by light
depends on which of the following?
I. Intensity of the light
II. Frequency of the light
III. Nature of the photoelectric surface
(A) I only
(B) II only (C) III only (D) II and III only
Speed depends on Kelectron = Ephoton – , which depends on
D
f (Ephoton = hf) and nature of photoelectric surface ()
6. If photons of light of frequency f have momentum p,
photons of light of frequency 2f will have a momentum of
(A) 2p
(B) ½p
(C) p
(D) 4p
p = h/ and  = c/f  p = hf/c (p  f)
A
If f is doubled than p is also doubled
7. Light of a particular wavelength is incident on a metal
surface, and electrons are emitted from the surface as a
result. To produce more electrons per unit time but with
less kinetic energy per electron, one should
(A) Increase the intensity and decrease the wavelength.
(B) Increase the intensity and the wavelength.
(C) Decrease the intensity and the wavelength.
(D) Decrease the intensity and increase the wavelength.
More electrons = more intense light; reduced kinetic
B
energy = reduced frequency or increased wavelength
8. If the momentum of an electron doubles, its de Broglie
wavelength is multiplied by a factor of
(A) ¼
(B) ½
(C) 1
(D) 2
p = h/  p  1/
B
If p is doubled, then  has to be half
Questions 9-14 The diagram shows the energy levels for
hydrogen gas.
214
9.
What is the energy, in eV, of a photon emitted by an
electron as it moves from the n = 6 to the n = 2?
(A) 0.38 eV (B) 3.02 eV (C) 3.40 eV (D) 13.60 eV
E6 = -0.38 eV and E2 = -3.40 eV
B
E = E2 – E6 = -3.40 eV – (-0.38 eV) = -3.02 eV
10. The energy of the photon (in J) is closest to
(A) 6.1 x 10-20 J
(B) 4.8 x 10-19 J
(C) 5.4 x 10-19 J
(D) 2.2 x 10-18 J
B 3.02 eV x 1.6 x 10-19 J/1 eV = 4.8 x 10-19 J
11. What is the frequency of the emitted photon?
(A) 9.2 x 1013 s-1
(B) 7.3 x 1014 s-1
14
-1
(C) 8.2 x 10 s
(D) 3.3 x 1015 s-1
E = hf
B
f = 4.83 x 10-19 J/6.63 x 10-34 J•s = 0.7 x 1015 s-1
12. What is the wavelength of the emitted photon?
(A) 4.1 x 10-7 m
(B) 3.3 x 10-7 m
(C) 5.4 x 10-7 m
(D) 5.0 x 10-7 m
c = f
A
 = c/f = 3 x 108 m/s/7.3 x 1014 s-1 = 0.4 x 10-6 m
13. What is the minimum amount of energy needed to ionized
an electron that is initially in the n = 6 energy level?
(A) 13.22 eV
(B) 5.12 eV
(C) 0.38 eV
(D) 0.16 eV
Ionization occurs when n = 
C
E = E – E6 = 0 – (-0.38 J) = 0.38 J
14. A photon having energy of 9.4 eV strikes a hydrogen atom
in the ground state. Why is the photon not absorbed by
the hydrogen atom?
(A) The atom's orbital electron is moving too fast
(B) The photon striking the atom is moving too fast.
(C) The photon's energy is too small.
(D) The photon is being repelled by electrostatic force.
9.4 eV is not enough energy to change n
C
(the minimum is -3.40 eV – (-13.60 eV) = 10.2 eV)
15. Electrons with energy E have a de Broglie wavelength of
approximately . In order to obtain electrons whose de
Broglie wavelength is 2, what energy is required?
(A) ¼E
(B) ½E
(C) 2E
(D) 4E
E = p2/2m and p = h/E = h2/2m2  E  1/2
A
If  is doubled then E is ¼
Questions 16-18 Use the graphs to answer the questions.
(A)
(B)
(C)
(D)
19. What characteristic determines energy carried by a photon?
(A) amplitude
(B) phase
(C) frequency
(D) velocity
C E = hf  frequency determines amount of energy
20. Which color is associated with the greatest energy change?
(A) blue
(B) red
(C) green (D) violet
E = hf  greatest energy requires highest frequency—
D
violet has the highest frequency
Practice Free Response
1. The fusion of a proton (11p = 1.007276 u) and neutron
(10n = 1.008665 u) produces deuterium, 21H (2.013553 u).
a. Write a nuclear equation.
1
1p
b.
Calculate the change in mass in u.
m = mH-2 – mp – mn
2.013553 – 1.007276 + 1.008665 = -0.002388 u
c.
Calculate the change in mass in kg (1 u = 1.66 x 10-27 kg)
-0.002388 u x 1.66 x 10-27 kg/u = -3.96 x 10-30 kg
d.
Calculate the energy in joules released.
E = mc2 = (-3.96 x 10-30 kg)(3 x 108 m/s)2 = -3.57 x 10-13 J
e.
If the deuterium absorbs all of the energy from part d
in the form of kinetic energy, what is the speed of the
deuterium particle?
m = 2.013553 u x 1.66E-27 kg/u = 3.34E-27 kg
K = ½mv2 3.57E-13 = ½(3.34E-27)v2v = 1.46E7 m/s
2.
The diagram shows the lowest four energy levels of an
atom. An electron in n = 4 state makes a transition to n = 2.
Determine
a. the energy of the emitted photon in eV.
E4 = -B/n2 = -(54.4 eV)/42 = -3.4 eV
E = E2 – E4 = -13.6 eV – (-3.4 eV) = -10.2 eV
b.
the wavelength of the emitted photon.
E = 1240 eV•nm/nm
nm = 1240 eV•nm/10.2 eV = 122 nm
c.
16. Which graph shows the de Broglie wavelength of a particle
versus the linear momentum?
p = h/  p 1/ (when p is large then  is small and visa
B
versa  B)
17. Which graph shows the maximum kinetic energy of the
emitted electrons versus the frequency of the light?
Kelectron = hfphoton – , which is the equation for a straight
A
line with a negative y-intercept  A
18. Which graph shows the total photoelectric current versus
the intensity of the light for a fixed frequency above the
cutoff frequency?
Intensity (# of photons)  current (# of electrons)—
D
intensity and current increase together
Questions 19-20 The spectrum of visible light emitted during
transitions in excited hydrogen atoms is composed of blue,
green, red, and violet lines
+ 10n  21H
the momentum of the emitted photon.
p = h/
p = (6.63 x 10-34 J•s)/(122 x 10-9 m) = 5.44 x 10-27 kg•m/s
The photon is incident on silver, which emits a photoelectron
(m = 9.11 x 10-31 kg) with wavelength  = 5.2 x 10-10 m.
Determine
d. the momentum of the photoelectron in kg•m/s.
p = h/ = 6.63 x 10-34 J•s/5.2 x 10-10 m
p = 1.3 x 10-24 kg•m/s
e.
the kinetic energy of the photoelectron in eV.
K = p2/2m = (1.3 x 10-24)2/2(9.11 x 10-31) = 8.9 x 10-19 J
8.9 x 10-19 J x 1 eV/1.6 x 10-19 J = 5.6 eV
f.
the work function of silver.
Kelectron = Ephoton – 
5.6 eV = 10.2 eV –    = 4.6 eV
Unit 6 Answers (Don't look until after you have tried the problem)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
D Strong nuclear forces > electric repulsion
It takes energy to remove an electron from neutral atom 
A separate parts have more total mass
C Negative charge attracts + alpha particle
A Positive charge attracts – beta particle
B Gamma is uncharged  not attracted to + or – charge
Nonspontaneous  energy (mass) is added to reactants
B form products
C Equal impulse (Ft) at separation  equal momentum
B Same momentum, but He has less mass  greater velocity
B mHevHe = mThvTh, vHe > vTh and v is squared in K  KHe > KTh
C 31H  0-1 + 32He
B 21084Po  42 + 20682Pb
C 146C + 0-1e  145B
C 10n + 168O  157N + 21H
C 90 years = 3 t½ (16 g  8 g  4 g  2 g)
A 16 g  8 g = 1 t½ (30 yr)
D 1  ½  ¼ (2 t½)  16 yr/2 = 8 yr
B Shorter t½ equals more radioactive
C  has no mass or charge  it can penetrate matter easily
Subatomic particle found in the nucleus
Combination of protons and neutrons
Atomic number = number of protons
Mass number = number of protons + neutrons
same Z value, but different A value
12.0 u x 1.66 x 10-27 kg/1 u = 1.99 x 10-26 kg
E = mc2 = (1.99 x 10-26 kg)(3 x 108 m/s)2 = 1.79 x 10-9 J
235
92
143
92
235
m = 2(mp) + 2(mn) – mHe
m = 2(1.007276) + 2(1.008665) – 4.001504 = 0.030378 u
0.030378 u x 1.66 x 10-27 kg/1u = 5.04 x 10-29 kg
E = mc2 = (5.04 x 10-29 kg)(3 x 108 m/s)2 = 4.54 x 10-12 J
4.54 x 10-12 J/4 = 1.13 x 10-12 J
4
4
2
low
2He
0
0
-1
medium
-1e

0
0
high
10
1
50
B + n  + 7Li
H + n  + 2H
Cr + n  + 51Cr
m = mHe + mTh - mU
m = 4.002602 u + 234.055381 u – 238.050784 u = 0.007199 u
0.007199 u x 1.66 x 10-27 kg/1 u = 1.20 x 10-29 kg
m > 0  reaction is not spontaneous.
E = mc2= (1.20 x 10-29 kg)(3 x 108 m/s)2 = 1.08 x 10-12 J
12 days = 3 t½  (1/8)1.776 g = 0.222 g
4.00 u x 1.66 x 10-27 kg/1 u = 6.64 x 10-27 kg
E = mc2= (6.64 x 10-27 kg)(3 x 108 m/s)2 = 5.98 x 10-10 J
226
86
140
86
226
m = 1mp + 2mn – mH-3
m = (1.007276) + 2(1.008665) – (3.015500) = 0.009106 u
m = 0.009106 u x 1.66 x 10-27 kg/1 u = 1.51 x 10-29 kg
BE = mc2 = (1.51 x 10-29 kg)(3 x 108 m/s)2 = 1.36 x 10-12 J
BE/A = 1.36 x 10-12 J/3 = 4.53 x 10-13 J/nucleon
27
9
Al +  1n + 30P
Be +  n + 12C
226
222
4
88Ra 
86Rn + 2He
221.97036 + 4.001504 – 225.97709 = -0.005226 u
-0.005226 u x 1.66 x 10-27 kg/1 u = -8.675 x 10-30 kg
E = mc2= (8.678 x 10-30 kg)(3 x 108 m/s)2 = 7.81 x 10-13 J
mRavRa = mv (222)(4) = (4)(222)
K/KRa = ½mv2/½mRavRa2 = (4)(222)2/(222)(4)2 = 222/4 = 55.5
E = ½mv2
(.98)(7.81 x 10-13 J) = ½(4 x 1.67 x 10-27 kg)v2 v = 1.51 x 107 m/s
 = h/mv = 6.63 x 10-34 J•s/(4 x 1.67 x 10-27 kg)(1.51 x 107 m/s)
 = 6.57 x 10-15 m
The loss of 7/8 activity means that 1/8 remains It takes 3 halflives to reach 1/8 (1  1/2  1/4  1/8) 3 x 14 = 42 days
f

E
m
p
I
violet
red
violet
violet
violet
?
8
-10
18 -1
c = f  3 x 10 m/s = f(1.54 x 10 m)f = 1.95 x 10 s
E = hf = (6.63 x 10-34 J•s)(1.95 x 1018 s-1) = 1.29 x 10-15 J
E = mc2  1.29 x 10-15 J = m(3 x 108 m/s)2 m = 1.44 x 10-32 kg
p = h/ = (6.63 x 10-34 J•s)/(1.54 x 10-10 m) = 4.31 x 10-24 kg•m/s
E = mc2= 2(1.67 x 10-27 kg)(3.0 x 108 m/s)2 = 3.0 x 10-10 J
E = hf  f = E/h = 3.0 x 10-10 J/6.63 x 10-34 J•s = 4.5 x 1023 s-1
c = f   = c/f = 3 x 108 m/s/4.5 x 1023 s-1 = 6.7 x 10-16 m
p = h/ = (6.63 x 10-34 J•s)/(6.7 x 10-16 m) = 1.0 x 10-18 kg•m/s
B E – E1 = 0 – (-13.6 eV/12) = 13.6 eV
The energy gap between 1 and 2 is ¾ of the total energy
D between ground state and ionization
would have the least energy  the electron would come
A Red
from the energy level closest to 2
D 4  3, 4  2, 4  1, 3  2, 3  1, 2  1
B Greater wavelength equals less energy (E = hc/)  B
A Greater frequency equals more energy (E = hf) A
D red < orange < yellow < green < blue < violet blue
light has more energy than 400-nm light; the extra
C 300-nm
energy becomes electron kinetic energy
44
45
46
47
48
49
50
51
52
53
54
55
56
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1
2
A Brighter (intense) has more photons, but not more energetic
B  = h/mv    1/v  slower = longer wavelength
B  = h/mv    1/m  lighter = longer wavelength
C  = h/p    1/p same p = same wavelength
c = f  3 x 108 m/s = f(750 x 10-9 m)f = 4.0 x 1014 s-1
E = hf = (6.63 x 10-34 J•s)(4.0 x 1014 s-1) = 2.7 x 10-19 J
E = mc2  2.7 x 10-19 J = m(3 x 108 m/s)2 m = 2.9 x 10-36 kg
p = h/ = (6.63 x 10-34 J•s)/(750 x 10-9 m) = 8.8 x 10-28 kg•m/s
E2 = -B/n2 = -13.6 eV/22 = -3.4 eV
E3 = -B/n2 = -13.6 eV/32 = -1.5 eV
E = -1.5 eV – (-3.4 eV) = 1.9 eV
Ephoton = (1240 eV•nm)/nm  1.9 eV = 1240/nm  nm = 653 nm
43
42
41
32
31
21
-12 eV
-32 eV
-140 eV
-20 eV
-128 eV
-108 eV
20 eV, 128 eV, 108 eV
E1 = -13.6 eV/12 = -13.6 eV, E2 = -13.6 eV/22 = -3.4 eV
E = En-high – En-low = -3.4 – (-13.6) = 10.2 eV
Ephoton = 1240/nm  nm = 1240/10.2 = 122 nm
E = E – En = 0 – (-B)/n2 = 13.6 eV/22 = 3.4 eV
nm = 1240 eV•nm /eV = 1240 eV•nm /3.10 eV = 400 nm
E = 1240/nm = 1240 eV•nm/240 nm = 5.17 eV
K = Ephoton -  = 5.17 eV – 2.40 eV = 2.77 eV
Vstop = Kelectron = 1.64 eV
Ephoton = 1240/nm = 1240 eV•nm/230 nm = 5.39 eV
 = Ephoton – K = 5.39 – 1.64 = 3.75 eV
E = 1240 eV•nm/500 nm = 2.48 eV
K = Ephoton - = 2.48 eV – 2.28 eV = 0.20 eV
 = 1240 eV•nm/nm = 1240 eV•nm/2.28 nm = 544 nm
0 eV, the energy of the photon is less than 
500 nm x 10-9 m/1 nm = 500 x 10-9 m
p = h/ = 6.63E-34/500E-9
p = h/ = 6.63E-34/500E-9
p = 1.3E-27kg•m/s
p = 1.3E-27 kg•m/s
m = p/c = 1.3E-27/3E8= 4.4E-36 kg 9.1E-31 kg
v = p/m = 1.3E-27/9.1E-31
3E8 m/s
v = 1.4E3 m/s
f = c/ = 3E8/500E-9 = 6E14
f = v/ = 1.4E3/500E-9= 2.8E9
2
-31
3 2
-34
14
-19
E = hf = (6.63E )(6E ) = 4E J E = ½mv-25= ½(9.1E )(1.4E )
E = 8.9E J
4E-19 J x 1 eV/1.6E-19 J = 2.5 eV 8.9E-25 J x 1/1.6E-19 J = 5.6E-6 eV
Practice Multiple Choice (No calculator)
C Spontaneous loses energy (mass)  MX > MY + MZ
214
0
x
D
82Pb  -1 + yZ: 214 = 0 + x  x = 214, 82 = -1 + y  y = 83
D Taking apart takes energy  mBE is added to the reactants
B t½ = 30 min and 3 half lives: 4000  2,000  1000  500
depends on Kelectron = Ephoton – , which depends on f
D Speed
(Ephoton = hf) and nature of photoelectric surface ()
A p = h/ and  = c/f  p = hf/c  f: 2 x f = 2 x p
B More electrons = more intense
reduce K = reduce f or increase 
B p = h/  p  1/: 2 x p = ½ 
E6 = -0.38 eV and E2 = -3.40 eV
B E
= E2 – E6 = -3.40 eV – (-0.38 eV) = -3.02 eV
B 3.02 eV x 1.6 x 10-19 J/1 eV = 4.8 x 10-19 J
B E = hf  f = 4.83 x 10-19 J/6.63 x 10-34 J•s = 0.7 x 1015 s-1
A c = f = c/f = 3 x 108 m/s/7.3 x 1014 s-1 = 0.4 x 10-6 m
Ionization occurs when n = 
C E = E – E = 0 – (-0.38 J) = 0.38 J

6
eV is not enough energy to change n
C 9.4
(the minimum is -3.40 eV – (-13.60 eV) = 10.2 eV)
E = p2/2m and p = h/E = h2/2m2  E  1/2
A If  is doubled then E is ¼
B p = h/  p 1/ (inverse relation  B)
A Kelectron = hfphoton – , straight line with negative y-intercept  A
Intensity (# of photons)  current (# of electrons)— intensity
D and current increase together
C E = hf  frequency determines amount of energy
= hf  greatest energy requires highest frequency—violet
D E
has the highest frequency
Practice Free Response
1
1
2
1p + 0n  1H
m = mH-2 – mp – mn
2.013553 – 1.007276 + 1.008665 = -0.002388 u
-0.002388 u x 1.66 x 10-27 kg/u = -3.96 x 10-30 kg
E = mc2 = (-3.96 x 10-30 kg)(3 x 108 m/s)2 = -3.57 x 10-13 J
m = 2.013553 u x 1.66E-27 kg/u = 3.34E-27 kg
K = ½mv2 3.57E-13 = ½(3.34E-27)v2v = 1.46E7 m/s
E4 = -B/n2 = -(54.4 eV)/42 = -3.4 eV
E = E2 – E4 = -13.6 eV – (-3.4 eV) = -10.2 eV
E = 1240 eV•nm/nm  nm = 1240 eV•nm/10.2 eV = 122 nm
p = h/ = (6.63 x 10-34 J•s)/(122 x 10-9 m) = 5.44 x 10-27 kg•m/s
p = h/ = 6.63 x 10-34 J•s/5.2 x 10-10 m = 1.3 x 10-24 kg•m/s
K = p2/2m = (1.3 x 10-24)2/2(9.11 x 10-31) = 8.9 x 10-19 J
8.9 x 10-19 J x 1 eV/1.6 x 10-19 J = 5.6 eV
Kelectron = Ephoton –   5.6 eV = 10.2 eV –    = 4.6 eV