Transcript The Nucleus: A Chemist’s View - University of California

The Nucleus: A Chemist’s
View
Chapter 20
E-mail: [email protected]
Web-site: http://clas.sa.ucsb.edu/staff/terri/
The Nucleus: A Chemist’s View
1. Write an equation for each of the following:
a. 68Ga undergoing electron capture
b. 62Cu undergoing positron emission
c. 212Fr undergoing alpha decay
d. 129Sb undergoing beta decay
The Nucleus: A Chemist’s View
Type of
Decay
Nuclear
Problem
Nuclear
Transformation
Emitted
Particle
Danger
Level
Alpha (α)
Nucleus is too
big (Z > 84)
Lose 2 11𝑝 and 2 10𝑛
A–4 and Z–2
4𝐻𝑒
Low
Beta (β–)
Neutron rich or
1𝑛: 1𝑝 is too
0
1
high
Convert neutron into
proton and electron
1𝑛  1𝑝 + 0𝑒
0
1
−1
Z+1
0
−1
Positron
Emission
(β+)
Proton rich or
1𝑝: 1𝑛 is too
1
0
high
Convert proton into
neutron and positron
1𝑝  1𝑛 + 0𝑒
1
0
+1
Z–1
0
+1
Electron
Capture
(EC)
Proton rich or
1𝑝: 1𝑛 is too
1
0
high
Convert proton into
neutron by absorbing
an electron
0
1
1
−1𝑒 + 1𝑝  0𝑛
Z–1
Gamma
Emission(γ)
Nucleus has too
much energy
Lose energy in the
form of gamma
photons
2
𝑒 or β–
Medium
𝑒 or β+
High
None
Low
0
0γ
or
gamma
photon
High
Example
230𝑇ℎ
90
42𝐻𝑒 + 226
88𝑅𝑎
14𝐶 
6
𝑒 + 147𝑁
0
−1
𝐵𝑖 +10𝑒 + 205
82𝑃𝑏
205
83
205
83
𝐵𝑖 + +10𝑒  205
82𝑃𝑏
98
43
98
𝐵𝑖 *  43
𝐵𝑖 + 00γ
The Nucleus: A Chemist’s View
2. Supply the missing particle for each of the following nuclear
reactions:
a. 73Ga  73Ge + ?
b. 192 Pt  188Os + ?
c. 205Bi  205Pb + ?
d. 241Cm + ?  241Am
The Nucleus: A Chemist’s View
3. What is the daughter nucleus if 242U underwent a decay series
producing 4 alpha and 3 beta particles?
The radioactive isotope 247Bk decays by an alpha and beta series ending
in 207Pb. How many alpha and beta particles were emitted in the
series?
The Nucleus: A Chemist’s View
4. The only stable isotope of fluorine is fluorine-19. Predict possible
modes of decay for fluorine-21 and fluorine-18.
The Nucleus: A Chemist’s View
5. The first atomic explosion was detonated on July16, 1945. What
fraction of the strontium-90 (t1/2 = 28.8 yr) will remain as of July
16, 2014?
The Nucleus: A Chemist’s View
6. A freshly isolated sample of 90Y was found to have an activity of
9.8×105 disintegrations per minute at 1:00 pm on December 3,
2000. At 2:15 pm on December 17, 2000, its activity was
redetermined and found to be 2.6×104 disintegrations per minute.
Calculate the half-life of 90Y.
The Nucleus: A Chemist’s View
7. Phosphorus-32 is a commonly used radioactive nuclide in
biochemical research. The half-life of phosphorus-32 is 14.3
days. What mass of phosphorus-32 is left from an original
sample of 175 mg of Na332PO4 after 35.0 days? Assume the
atomic mass of 32P is 32.0.
The Nucleus: A Chemist’s View
8. A rock contains 0.688 mg of 206Pb for every 1.000 mg of 238U
present. Calculate the age of the rock (t1/2 = 4.5x109 yr).
Assume that no lead was present in the original rock.
The Nucleus: A Chemist’s View
9. The sun radiates 3.9 x 1023 J of energy into space every second.
What is the rate at which mass is lost from the sun?
The Nucleus: A Chemist’s View
10. The most stable nucleus in terms of binding energy per nucleon is
56Fe. If the atomic mass of 56Fe is 55.9349 amu, calculate the
binding energy per nucleon for 56Fe. (neutron = 1.67493 x 10-27
kg, proton = 1.67262 x 10-27 kg, electron = 9.10939 x 10-31 kg)
The Nucleus: A Chemist’s View
11. A positron and an electron annihilate each other upon colliding,
thereby producing energy in the form of 2 photons. Calculate
the wavelength of the light produced.
(mass of an electron = 9.10939 x10-31kg)
The Nucleus: A Chemist’s View
1. Write an equation for each of the following:
a. 68Ga undergoing electron capture
68Ga + 0𝑒  68Zn
−1
b. 62Cu undergoing positron emission
62Cu  0𝑒 + 62Ni
−1
c. 212Fr undergoing alpha decay
212Fr  4𝐻𝑒 + 208At
2
d. 129Sb undergoing beta decay
129Sb  0𝑒 + 52Te
−1
The Nucleus: A Chemist’s View
2. Supply the missing particle for each of the following nuclear
reactions:
0
a. 73Ga  73Ge + ?
−1𝑒 ⇒ beta particle
4
b. 192 Pt  188Os + ?
2𝐻𝑒 ⇒ alpha particle
0
c. 205Bi  205Pb + ?
+1𝑒 ⇒ positron
0
d. 241Cm + ?  241Am
−1𝑒 ⇒ electron
The Nucleus: A Chemist’s View
3. What is the daughter nucleus if 242U underwent a decay series
producing 4 alpha and 3 beta particles?
242U  4 4𝐻𝑒 + 3 0𝑒 + 226Fr
2
−1
The radioactive isotope 247Bk decays by an alpha and beta series ending
in 207Pb. How many alpha and beta particles were emitted in the
series?
247Bk  207Pb + X 4𝐻𝑒 + Y 0𝑒
2
−1
247 = 207 + 4X
X = 10
97 = 82 + 2(10) –Y
Y=5
The Nucleus: A Chemist’s View
4. The only stable isotope of fluorine is fluorine-19. Predict possible
modes of decay for fluorine-21 and fluorine-18.
If F-19 is stable then F-21 is neutron rich and will undergo beta
decay – whereas F-18 is proton rich an will either undergo positron
emission or electron capture
The Nucleus: A Chemist’s View
5. The first atomic explosion was detonated on July16, 1945. What
fraction of the strontium-90 (t1/2 = 28.8 yr) will remain as of July
16, 2014?
All radioactive decay is first order kinetics ⇒ ln[A]= - kt + ln[A]o
and t1/2 = 0.693/k ⇒ the fraction remaining =
[𝐴]
𝐴𝑜
[𝐴]
𝐴𝑜
= 𝑒 −𝑘𝑡 or
[𝐴]
𝐴𝑜
[𝐴]
𝐴𝑜
= 𝑒 −0.693𝑡/t1/2
= 𝑒 −0.693(69𝑦𝑟)/(28.8𝑦𝑟) = 0.2 or 20% remains
The Nucleus: A Chemist’s View
6. A freshly isolated sample of 90Y was found to have an activity of 9.8×105
disintegrations per minute at 1:00 pm on December 3, 2000. At 2:15 pm on
December 17, 2000, its activity was re-determined and found to be 2.6×104
disintegrations per minute. Calculate the half-life of 90Y in hrs.
t1/2 = 0.693/k
you can get k from the integrated rate law ⇒ ln[A] = -kt + ln[A]o
since you’re not given [A] and [A]o you can substitute in ⇒ rate = k[A]
ln(rate/k) = -kt + ln(rateo/k)
t = 14days + 1hr + 15min = 337.25hr
𝑟𝑎𝑡𝑒/𝑘
k=
𝑙𝑛𝑟𝑎𝑡𝑒 /𝑘
𝑜
−𝑡
⇒ t1/2 =
−0.693𝑡
𝑟𝑎𝑡𝑒/𝑘
𝑙𝑛𝑟𝑎𝑡𝑒 /𝑘
𝑜
=
−0.693(337.25ℎ𝑟)
6.6𝑥104 𝑑𝑝𝑚
𝑙𝑛9.8𝑥105 𝑑𝑝𝑚
= 64.4hr
The Nucleus: A Chemist’s View
7. Phosphorus-32 is a commonly used radioactive nuclide in biochemical
research. The half-life of phosphorus-32 is 14.3 days. What mass of
phosphorus-32 is left from an original sample of 175 mg of Na332PO4 after
35.0 days? Assume the atomic mass of 32P is 32.0.
32
𝑃
32 𝑔𝑚𝑜𝑙
In the original sample ⇒ 175 mg Na332PO4 164.97𝑔
𝑚𝑜𝑙
using ln[A] = -kt + ln[A]o and t1/2 = 0.693/k
ln[A] =
−0.693𝑡
𝑡1 2
ln[A] =
−0.693(35 𝑑𝑎𝑦𝑠)
(14.3 𝑑𝑎𝑦𝑠)
+ ln[A]o
/
[A] = 6.22 mg
+ ln(33.95 mg)
32
𝑁𝑎3 𝑃𝑂4
= 33.95 mg 32P
The Nucleus: A Chemist’s View
8. A rock contains 0.688 mg of 206Pb for every 1.000 mg of 238U
present. Calculate the age of the rock (t1/2 = 4.5x109 yr).
Assume that no lead was present in the original rock.
If the final rock has 0.688 mg of 206Pb and 1 mg of 238U then the original rock
must have had 1.688 mg of 238U. Using ln[A] = -kt + ln[A]0 and t1/2 = 0.693/t
[1𝑚𝑔]
[𝐴]
t=
𝑙𝑛 𝐴
0
0.693
− 𝑡
1 2
/
=
𝑙𝑛 1.688 𝑚𝑔
0.693
−4.5𝑥109 𝑦𝑟
= 3.4x109 yr
The Nucleus: A Chemist’s View
9. The sun radiates 3.9 x 1023 J of energy into space every second.
What is the rate at which mass is lost from the sun?
Fusion occurs on the sun – fusion is the combining of 2 small
nuclei into 1 nucleus – a small amount of matter is converted to E
due to making the strong forces in the nucleus
ΔE = Δmc2
-3.9x1023 J =Δm(3x108 m/s)2
Δm = -4.6x106 kg
The Nucleus: A Chemist’s View
10. The most stable nucleus in terms of binding energy per nucleon is
56Fe. If the atomic mass of 56Fe is 55.9349 amu, calculate the
binding energy per nucleon for 56Fe. (neutron = 1.67493 x 10-27 kg,
proton = 1.67262 x 10-27 kg, electron = 9.10939 x 10-31 kg)
Binding energy ⇒ the amount of energy required to break apart a nucleus into
protons and neutrons ⇒ 56Fe26+  26 11𝑝 + 30 10𝑛
ΔE = Δmc2
0.001𝑘𝑔
ΔE = [26(1.67262 x 10-27 kg) + 30(1.67493 x 10-27 kg) –(55.9349 amu
6.022𝑥1023 𝑎𝑚𝑢
-31
8
2
- 26(9.10939 x 10 kg))](3x10 m/s)
7.879x10−11J/nucleus
ΔE =
= 1.407x10-12 J/nucleon
56 𝑛𝑢𝑐𝑙𝑒𝑜𝑛𝑠/𝑛𝑢𝑐𝑙𝑒𝑢𝑠
The Nucleus: A Chemist’s View
11. A positron and an electron annihilate each other upon colliding,
thereby producing energy in the form of 2 photons. Calculate
the wavelength of the light produced.
(mass of an electron = 9.10939 x10-31kg)
Annihilation ⇒ Conversion of all mass into energy when matter and
antimatter collide −10𝑒 + +10𝑒  2 00γ
Since an electron and a positron have the same mass then the energy of the 2
photons is also the same
ΔE = Δmc2
ΔE = (0 kg - 9.10939x10-31kg)(3x108 m/s)2
ΔE = - 8.2x10-14J
λ = hc/E = (6.626x10-34J)(3x108 m/s)/(8.2x10-14J) ⇒ λ = 2.42x10-12m