Physics 204 – Section 9
QUIZ 11
5 May 2014
Name:
60 Co,
27
1. (10 total points) You have a 0.2 g sample of
59.933 819 u. It’s half-life is 5.271 yr.
which has an atomic mass of
(a) (3 points) Find the number of 60
27 Co atoms in your sample.
Solution: Cobalt-60 has an atomic mass of 59.933 819 u, so 1 mol (6.022 ⇥
60 Co atoms) has a mass of 59.933 819 g. Then
1023 27
N = 0.2 g
1 mol
6.022 ⇥ 1023
= 2.009 55 ⇥ 1021 .
59.933 819 g
1 mol
(1)
(b) (2 points) Find the decay constant l for 60
27 Co.
Solution: Half-life is related to the decay constant by
ln 2
,
l
(2)
ln 2
ln 2
1 yr
1 day 1 h
=
T1/2
5.271 yr 365.25 days 24 h 3600 s
(3)
T1/2 =
so
l =
= 4.167 05 ⇥ 10
9
1
s
= 0.131 502 yr
1
.
(4)
(c) (5 points) Find the activity of your sample.
Solution: Activity
A = lN
⇣
= 4.167 05 ⇥ 10
9
s
1
⌘⇣
2.009 55 ⇥ 1021
= 8.3739 ⇥ 1012 Bq = 226.321 Ci
⌘
= 2.6426 ⇥ 1020 decays/year.
NA = 6.022 ⇥ 1023 mol
1
T1/2 =
0.693
l
A = lN
A = A0 e
(5)
(6)
lt
N = N0 e
lt
Physics 204 – Section 11
QUIZ 11
5 May 2014
Name:
1. (10 total points) You start with a sample of 222
86 Rn (radon-222) that has an activity of
308 Ci. Radon-222 has a half-life of 3.825 days.
(a) (2 points) Find the decay constant l.
Solution: The half-life is related to the decay constant by
T1/2 =
ln 2
,
l
(1)
so the decay constant
l=
ln 2
ln 2
=
= 0.181215 days
T1/2
3.825 days
1
= 2.0974 ⇥ 10
6
s
1
.
(2)
(b) (4 points) After one day, what percentage of the 222
86 Rn is left?
Solution: The number of 222
86 Rn atoms after time t is
N = N0 e
lt
.
(3)
Therefore, the percentage remaining after one day is
N
=e
N0
lt
=e
(0.181215 days
1 )(1 day)
= 83.4256%.
(4)
(c) (4 points) After one day, what is the activity of your sample?
Solution: The activity decays with time as the radioactive atoms decay.
Therefore, the activity
DN =
lNDt
lt
= 308 Cie
N = N0 e
(0.181215 days
1 )(1 day)
= 256.951 Ci = 9.50718 ⇥ 1012 Bq.
(5)
The activity has decreased by 83.4% just like the number of 222
86 Rn atoms.
A = A0 e
lt
A = lN
A = A0 e
lt
T1/2 =
0.693
l
Physics 204 – Section 13
QUIZ 11
5 May 2014
Name:
1. (10 total points) .
(a) (3 points) Find the mass defect Dm of
of 12 u.
12 C
6
(carbon-12) if the isotope has a mass
Solution: Carbon-12 has 6 protons, 6 neutrons, and 6 electrons. Therefore,
its mass defect is
Dm = (6m p + 6mn + 6me )
m12 C
6
= [6(1.007 276 u) + 6(1.008 665 u) + 6(5.485 799 ⇥ 10
4
u)]
= 0.098 938 u.
12 u
(1)
(b) (2 points) Find the binding energy of 12
6 C.
Solution: When 1 u of mass is converted into energy, it produces 931.5 MeV.
Therefore, the binding energy of 12
6 C is
E = 0.098 938 u
931.5 MeV
= 92.1607 MeV.
1u
(c) (3 points) Find the mass defect Dm of
of 14.003 241 u.
14 C
6
(2)
(carbon-14) if the isotope has a mass
Solution: Carbon-14 has 6 protons, 8 neutrons, and 6 electrons. Therefore,
its mass defect is
Dm = (6m p + 8mn + 6me )
m12 C
6
= [6(1.007 276 u) + 8(1.008 665 u) + 6(5.485 799 ⇥ 10
= 0.113 026 u.
4
u)]
14.003 241 u
(3)
(d) (2 points) Find the binding energy of 14
6 C.
m p = 1.007 276 u
mn = 1.008 665 u
me = 5.485 799 ⇥ 10
4
u
1 u ! 931.5 MeV
Physics 204 – Section 13
QUIZ 11
5 May 2014
Solution: The binding energy is
E = 0.113 026 u
m p = 1.007 276 u
mn = 1.008 665 u
931.5 MeV
= 105.284 MeV
1u
me = 5.485 799 ⇥ 10
4
u
(4)
1 u ! 931.5 MeV
Physics 204 – Section 20
QUIZ 11
5 May 2014
Name:
1
12
4
1. (10 total points) The sun creates carbon-12 by the reaction 15
7 N + 1 H ! 6 C + 2 He.
The atomic masses for these isotopes are: 1.007 825 u for 11 H, 4.002 603 u for 42 He,
12
15.000 108 u for 15
7 N, and 12 u for 6 C. 1 u of mass contains 931.5 MeV of energy.
1
(a) (3 points) Find the total mass of the initial atoms 15
7 N + 1 H.
Solution: The initial mass
m15 N + m1 H = 15.000 108 u + 1.007 825 u = 16.007 933 u.
7
1
(1)
4
(b) (3 points) Find the total mass of the final atoms 12
6 C + 2 He.
Solution: The final mass
m12 C + m4 He = 12 u + 4.002 603 u = 16.002 603 u.
6
(2)
2
(c) (4 points) Find the energy released by this reaction.
Solution: The mass decreases by
Dm = 16.007 933 u
16.002 603 u = 0.005 33 u.
(3)
Converting 1 u of mass to energy yields 931.5 MeV. Therefore, the energy
released by this reaction is
E = 0.005 33 u
m p = 1.007 276 u
931.5 MeV
= 4.964 892 MeV.
1u
mn = 1.008 665 u
me = 5.485 799 ⇥ 10
(4)
4
u