Practice Problems
20. a.
12.0 u x 1.66 x 10-27 kg/1 u = 1.99 x 10-26 kg
1.
D
b.
Strong nuclear forces > electric repulsion
E = mc2 = (1.99 x 10-26 kg)(3 x 108 m/s)2 = 1.79 x 10-9 J
2.
A
It takes energy to remove an electron from neutral atom
 separate parts have more total mass
21.
3.
C
Positive charge attracts – beta particle
5.
B
Number of
neutrons
Z value
A value
235
92
143
92
235
22. a. (1)
m = 2(mp) + 2(mn) – mHe
m = 2(1.007276) + 2(1.008665) – 4.001504 = 0.030378 u
(2)
Gamma is uncharged  not attracted to + or – charge
0.030378 u x 1.66 x 10-27 kg/1u = 5.04 x 10-29 kg
6.
b.
Nonspontaneous  energy (mass) is added to
B
reactants form products
7. Which has greater momentum, A—Th-234, or B—He-4?
(A) A
(B) B
(C) tie
C
Number of
protons
Negative charge attracts + alpha particle
4.
A
Number of
nucleons
Equal impulse (Ft) at separation  equal momentum
E = mc2 = (5.04 x 10-29 kg)(3 x 108 m/s)2 = 4.54 x 10-12 J
c.
4.54 x 10-12 J/4 = 1.13 x 10-12 J
23.
8.
B
Same momentum, but He has less mass  greater
velocity
mHevHe = mThvTh, but in kinetic energy, v is squared 
He has more kinetic energy
gamma
3
1H
 0-1 + 32He
10B
84Po
 2 +
4
14
6C
+ 0-1e  145B
1
0n
+ 168O  157N + 21H
90 years = 3 t½ (16 g  8 g  4 g  2 g)
0
0
high
+ n  + 7Li
1H
+ n  + 2H
u 4He = 4.002602 u
16 g  8 g = 1 t½ (30 yr)
26.
1  ½  ¼ (2 t½)  16 yr/2 = 8 yr
12 days = 3 t½
(1/8)1.776 g = 0.222 g
27. a.
50Cr
+ n  + 51Cr
234Th = 234.055381
4.00 u x 1.66 x 10-27 kg/1 u = 6.64 x 10-27 kg
Shorter t½ equals more radioactive
18.
C

c.
E = mc2
E = (1.20 x 10-29 kg)(3 x 108 m/s)2 = 1.08 x 10-12 J
17.
B
medium
m > 0  reaction is not spontaneous.
16.
D
-1
b.
15.
A
0
0.007199 u x 1.66 x 10-27 kg/1 u = 1.20 x 10-29 kg
14.
C
-1e
u
a. (1)
m = mHe + mTh - mU
m = 4.002602 u+234.055381 u–238.050784 u=0.007199 u
(2)
206 Pb
82
13.
C
Penetrating ability
low
238U = 238.050784
210
12.
C
Charge
2
25.
11.
B
Mass
4
24.
10.
C
0
beta
9.
B
Symbol
4 He
2
alpha
Gamma radiation is pure energy  with charge or mass,
28.
it can penetrate matter without much loss in energy
19.
nucleon
Subatomic particle found in the nucleus
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
b.
E = mc2
E = (6.64 x 10-27 kg)(3 x 108 m/s)2 = 5.98 x 10-10 J
Number of
nucleons
226
29.
m
(u)
m
(kg)
Number of
protons
86
Number of
neutrons
140
Z value
A value
86
226
m = 1mp + 2mn – mH-3
m = (1.007276) + 2(1.008665) – (3.015500)
m = 0.009106 u
m = 0.009106 u x 1.66 x 10-27 kg/1 u
m = 1.51 x 10-29 kg
BE = mc2 = (1.51 x 10-29 kg)(3 x 108 m/s)2
BE = 1.36 x 10-12 J
BE
4  3, 4  2, 4  1, 3  2, 3  1, 2  1
D
BE/A BE/A = 1.36 x 10-12 J/3 = 4.53 x 10-13 J/nucleon
40.
Greater wavelength equals less energy (E = hc/)  B
B
30.
27Al
+  1n + 30P
9Be
+  n + 12C
Greater frequency equals more energy (E = hf) A
A
31. a.
226
41.
88Ra
 22286Rn + 42He
42.
b.
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
c.
E = mc2
E = (8.678 x 10-30 kg)(3 x 108 m/s)2 = 7.81 x 10-13 J
d.
mRavRa = mv (222)(4) = (4)(222)
K = ½mv2 = (4)(222)2 = 222 = 55.5
KRa ½mRavRa2 (222)(4)2
4
e.
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.
 = 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.
The loss of 7/8 activity means that 1/8 remains = 3 t½
(1  1/2  1/4  1/8) 3 x 14 = 42 days
red < orange < yellow < green < blue < violet blue
D
43.
300-nm light has more energy than 400-nm light; the
extra energy becomes electron kinetic energy
C
44.
Brighter, more intense light, has more photons, but not
more energetic photons
A
45.
 = h/mv    1/v  slower = longer wavelength
B
46.
 = h/mv    1/m  lighter = longer wavelength
B
47.
 = h/p    1/p same p = same wavelength
C
48.
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
33.
yellow
violet
can't tell
frequency
wavelength
energy
relativistic mass
momentum
intensity
34.
c = f
3 x 108 m/s = f(1.54 x 10-10 m)f = 1.95 x 1018 s-1
E = hf
E
E = (6.63 x 10-34 J•s)(1.95 x 1018 s-1) = 1.29 x 10-15 J
E = mc2
m
1.29 x 10-15 J = m(3 x 108 m/s)2 m = 1.44 x 10-32 kg
p = h/
p
p = (6.63 x 10-34 J•s)/(1.54 x 10-10 m) = 4.31 x 10-24 kg•m/s
f
photon
E = mc2= 2(1.67 x 10-27 kg)(3.0 x 108 m/s)2
E = 3.0 x 10-10 J
E = hf  f = E/h
f
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 = h/ = (6.63 x 10-34 J•s)/(6.7 x 10-16 m)
p
p = 1.0 x 10-18 kg•m/s
36.
B
The energy gap between 1 and 2 is ¾ of the total
energy between ground state and ionization
38.
A
39.
n=4
-16
n=3
-36
n=2
Transition
E
n = 4  n = 3 -16 - -4 = -12 eV
n = 4  n = 2 -36 - -4 = -32 eV
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
n=1
n = 2  n = 1 -144 - -36 = -108 eV
20 eV, 128 eV, 108 eV
50.
nm = 1240 eV•nm /eV = 1240 eV•nm /3.10 eV = 400 nm
51. a.
E = 1240/nm = 1240 eV•nm/240 nm = 5.17 eV
b.
K = Ephoton -  = 5.17 eV – 2.40 eV = 2.77 eV
E – E1 = 0 – (-13.6 eV/12) = 13.6 eV
37.
D
-4
b.
35.
E
Ephoton = (1240 eV•nm)/nm
1.9 eV = (1240 eV•nm)/nm  nm = 653 nm
49. a.
Energy (eV)
red
Red would have the least energy  the electron would
come from the energy level closest to 2
52. a.
 = c/f = 3 x 108 m/s/7 x 1014 s-1 = 4.3 x 10-7 m (430 nm)
b.
Kelectron = y value when f = 7 x 1014 s-1 = 0.59 eV
c.
 = |y-intercept| = 2.3 eV
d.
11.
fthreshold = x-intercept = 5.6 x
1014
s-1
B
e.
 = c/f = 3 x 108 m/s/5.6 x 1014 s-1 = 5.4 x 10-7 m (540 nm)
 = 1240 eV/Ethreshold = 1240 eV•nm/2.3 eV = 540 nm
f.
Ephoton = 1240/nm = 1240 eV•nm/430 nm = 2.9 eV
Ephoton = Kelectron +  = 0.59 eV + 2.3 eV = 2.9 eV
g.
h = slope = [(0.59 – -2.3) eV/7 x 1014 s-1](1.6 x 10-19 J/eV)
h = 6.6 x 10-34 J•s
53.
12.
A

(m)
p
(kg•m/s)
m
(Kg)
v
(m/s)
f
(s-1)
E
(J)
E
(eV)
13.
D
B
500 nm x
10-9
electron
m/1 nm = 500 x
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
3E8
m/s
f = c/
f = 3E8/500E-9
f = 6E14
E = hf
E = (6.63E-34)(6E14)
E = 4E-19 J
4E-19 J x 1 eV
1.6E-19 J
E = 2.5 eV
10-9
m
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 = v/
f = 1.4E3/500E-9
f = 2.8E9
E = ½mv2
E = ½(9.1E-31)(1.4E3)2
E = 8.9E-25 J
8.9E-25 J x 1 eV
1.6E-19 J
-6
E = 5.6E eV
B
B
17.
A
C
C
A
6.
A
p = h/ and  = c/f  p = hf/c (p  f)
If f is doubled than p is also doubled
7.
B
More electrons = more intense light; reduced kinetic
energy = reduced frequency or increased wavelength
8.
B
p = h/  p  1/
If p is doubled, then  has to be half
9.
C
E = hf  frequency determines amount of energy
10.
D
E = hf  greatest energy requires highest frequency—
violet has the highest frequency
E = p2/2m and p = h/E = h2/2m2  E  1/2
If  is doubled then E is ¼
a.
1 p + 1 n  2 H
1
0
1
b.
m = mH-2 – mp – mn
2.013553 – 1.007276 – 1.008665 = -0.002388 u
c.
d.
E = mc2 = (-3.96 x 10-30 kg)(3 x 108 m/s)2 = -3.57 x 10-13 J
Taking a deuteron apart takes energy  mBE, is added
to the reactant side. md + mBE = mp + mn
Speed depends on Kelectron = Ephoton – , which depends
on f (Ephoton = hf) and nature of photoelectric surface ()
9.4 eV is not enough energy to change n
(the minimum is -3.40 eV – (-13.60 eV) = 10.2 eV)
-0.002388 u x 1.66 x 10-27 kg/u = -3.96 x 10-30 kg
 0-1 + xyZ
214 = 0 + x  x = 214, 82 = -1 + y  y = 83  21483Bi
4.
D
Ionization occurs when n = 
E = E – E6 = 0 – (-0.38 J) = 0.38 J
Practice Free Response
1.
3.
5.
c = f
 = c/f = 3 x 108 m/s/7.3 x 1014 s-1 = 0.4 x 10-6 m
20.
82Pb
The counts/minute decrease by half in 30 min (t½)  at
1:00 the count is 1,000 and at 1:30 the count is 500
E = hf
f = 4.83 x 10-19 J/6.63 x 10-34 J•s = 0.7 x 1015 s-1
19.
Spontaneous process loses energy (mass) in order to
reach a lower energy state  MX > MY + MZ
214
3.02 eV x 1.6 x 10-19 J/1 eV = 4.8 x 10-19 J
18.
2.
B
E6 = -0.38 eV and E2 = -3.40 eV
E = E2 – E6 = -3.40 eV – (-0.38 eV) = -3.02 eV
16.
Practice Multiple Choice
D
Intensity (# of photons)  current (# of electrons)—
intensity and current increase together
14.
1.
D
Kelectron = hfphoton – , which is the equation for a
straight line with a negative y-intercept  A
15.
photon
C
p = h/  p 1/ (when p is large then  is small and
visa versa  B)
2.
e.
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
a.
E4 = -B/n2 = -(54.4 eV)/42 = -3.4 eV
E = E2 – E4 = -13.6 eV – (-3.4 eV) = -10.2 eV
b.
E = 1240 eV•nm/nm
nm = 1240 eV•nm/10.2 eV = 122 nm
c.
p = h/
p = (6.63 x 10-34 J•s)/(122 x 10-9 m) = 5.44 x 10-27 kg•m/s
d.
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.
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.
Kelectron = Ephoton – 
5.6 eV = 10.2 eV –    = 4.6 eV