Transcript The Nucleus: A Chemist*s View
The Nucleus: A Chemist’s View
Chapter 20
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The Nucleus: A Chemist’s View
1. Supply the missing particle for each of the following nuclear reactions: a. 73 Ga 73 Ge + ?
b. 192 Pt 188 Os + ?
c. 205 Bi 205 Pb + ?
d. 241 Cm + ? 241 Am
The Nucleus: A Chemist’s View
The Nucleus: A Chemist’s View
2. Write an equation for each of the following: a. 68 Ga undergoing electron capture b. 62 Cu undergoing positron emission c. 212 Fr undergoing alpha decay d. 129 Sb undergoing beta decay
The Nucleus: A Chemist’s View
3. The radioactive isotope 247 Bk decays by an alpha and beta series ending in 207 Pb. How many alpha and beta particles were emitted in the series?
What is the daughter nucleus if 242 U underwent a decay series producing 4 alpha and 3 beta particles?
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 (t
1/2
16, 2014?
= 28.8 yr) will remain as of July
The Nucleus: A Chemist’s View
6. The sun radiates 3.9 x 10 23 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
7. A freshly isolated sample of 90 Y was found to have an activity of 9.8×10 5 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 2.6×10 4 disintegrations per minute. Calculate the half-life of 90 Y.
The Nucleus: A Chemist’s View
8. 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 Na 3 32 PO 4 atomic mass of 32 P is 32.0. after 35.0 days? Assume the
The Nucleus: A Chemist’s View
9. A rock contains 0.688 mg of 206 Pb for every 1.000 mg of 238 U present. Calculate the age of the rock (t
1/2
= 4.5x10
9 Assume that no lead was present in the original rock.
yr).
The Nucleus: A Chemist’s View
10. The most stable nucleus in terms of binding energy per nucleon is 56 Fe. If the atomic mass of 56 Fe is 55.9349 amu, calculate the binding energy per nucleon for 56 Fe. (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 -31 kg)
The Nucleus: A Chemist’s View
1. Supply the missing particle for each of the following nuclear reactions: a. b. c. d. 73 192 205 241 Ga Pt Bi 73 Ge + ?
188 205 Cm + ? Os + ?
Pb + ?
241 Am 0 −1 𝑒 4 2 𝐻𝑒 0 +1 𝑒 ⇒
alpha particle
⇒
positron
0 −1 𝑒 ⇒ ⇒
beta particle electron
The Nucleus: A Chemist’s View
2. Write an equation for each of the following: a. 68 Ga undergoing electron capture
68 Ga +
0 −1 𝑒
68 Zn
b. 62 Cu undergoing positron emission
62 Cu
0 −1 𝑒
+ 62 Ni
c. 212 Fr undergoing alpha decay
212 Fr
4 2 𝐻𝑒
+ 208 At
d. 129 Sb undergoing beta decay
129 Sb
0 −1 𝑒
+ 52 Te
The Nucleus: A Chemist’s View
3. The radioactive isotope 247 Bk decays by an alpha and beta series ending in 207 Pb. How many alpha and beta particles were emitted in the series?
247 Bk
207 Pb + X
4 2 𝐻𝑒
+ Y
0 −1 𝑒
247 = 207 + 4X X = 10 97 = 82 + 2(10) –Y Y = 5
What is the daughter nucleus if 242 U underwent a decay series producing 4 alpha and 3 beta particles?
242 U
4
4 2 𝐻𝑒
+ 3
0 −1 𝑒
+ 226 Fr
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 (t
1/2
16, 2014?
= 28.8 yr) will remain as of July
All radioactive decay is first order kinetics and t 1/2
⇒
ln[A]= - kt + ln[A] o = 0.693/k
⇒
the fraction remaining =
[𝐴] 𝐴 𝑜 [𝐴] 𝐴 [𝐴] 𝑜 𝐴 𝑜
=
𝑒 −𝑘𝑡 or [𝐴]
=
𝑒 −0.693𝑡/
t 1/2
𝐴 𝑜
=
𝑒 −0.693(69𝑦𝑟)/(28.8𝑦𝑟)
= 0.2 or 20% remains
The Nucleus: A Chemist’s View
6. The sun radiates 3.9 x 10 23 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 a strong force in the nucleus ΔE = Δmc 2
The Nucleus: A Chemist’s View
7. A freshly isolated sample of 90 Y was found to have an activity of 9.8×10 5 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 2.6×10 4 disintegrations per minute. Calculate the half-life of 90 Y in hrs.
t 1/2 = 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(rate o /k) t = 14days + 1hr + 15min = 337.25hr
k =
𝑙𝑛 𝑟𝑎𝑡𝑒/𝑘 𝑟𝑎𝑡𝑒 𝑜 /𝑘 −𝑡 ⇒
t 1/2
−0.693𝑡
=
𝑙𝑛 𝑟𝑎𝑡𝑒/𝑘 𝑟𝑎𝑡𝑒 𝑜 /𝑘
=
−0.693(337.25ℎ𝑟) 𝑙𝑛 6.6𝑥104 𝑑𝑝𝑚 9.8𝑥105 𝑑𝑝𝑚
= 64.4hr
The Nucleus: A Chemist’s View
8. 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 Na 3 32 PO 4 35.0 days? Assume the atomic mass of 32 P is 32.0. after
In the original sample
⇒
175 mg Na 3 32 PO 4
32 𝑔 164.97𝑔 𝑚𝑜𝑙 32 𝑃 𝑚𝑜𝑙 𝑁𝑎 3 32 𝑃𝑂 4
= 33.95 mg 32 P using ln[A] = -kt + ln[A] o and t 1/2 = 0.693/k ln[A] =
−0.693𝑡 𝑡 1 / 2
+ ln[A] ln[A] =
−0.693(35 𝑑𝑎𝑦𝑠) (14.3 𝑑𝑎𝑦𝑠)
o + ln(33.95 mg) [A] = 6.22 mg
The Nucleus: A Chemist’s View
9. A rock contains 0.688 mg of 206 Pb for every 1.000 mg of 238 U present. Calculate the age of the rock (t
1/2
= 4.5x10
9 Assume that no lead was present in the original rock.
yr).
If the final rock has 0.688 mg of 206 Pb and 1 mg of 238 U then the original rock must have had 1.688 mg of 238 U. Using ln[A] = -kt + ln[A] 0 and t 1/2 = 0.693/t t =
𝑙𝑛 [𝐴] 𝐴 0 − 0.693
𝑡 1 / 2
=
𝑙𝑛 [1𝑚𝑔] 1.688 𝑚𝑔 − 0.693
4.5𝑥109 𝑦𝑟
= 3.4x10
9 yr
The Nucleus: A Chemist’s View
10. The most stable nucleus in terms of binding energy per nucleon is 56 Fe. If the atomic mass of 56 Fe is 55.9349 amu, calculate the binding energy per nucleon for 56 Fe. (neutron = 1.67493 x 10 -27 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
⇒
56 Fe 26+
26
1 1 𝑝
+
30 1 0 𝑛
ΔE = Δmc 2
kg,
ΔE = [26(1.67262 x 10 -27 kg) + 30(1.67493 x 10 - 26(9.10939 x 10 -31 -27
0.001𝑘𝑔
kg) –(55.9349 amu
6.022𝑥1023 𝑎𝑚𝑢
kg)](3x10 8 m/s) 2 ΔE = 7.879x10−11J/nucleus
56 𝑛𝑢𝑐𝑙𝑒𝑜𝑛𝑠/𝑛𝑢𝑐𝑙𝑒𝑢𝑠
= 1.407x10
-12 J/nucleon
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 -31 kg)
Annihilation
⇒
Conversion of all mass into energy when matter and antimatter collide
0 −1 𝑒
+
0 +1 𝑒
2
0 0 γ
Since the mass of an electron and a positron are the same then the energy of the 2 photons are the same ΔE = Δmc 2 ΔE = (0 kg - 9.10939x10
-31 kg)(3x10 8 m/s) 2 ΔE = - 8.2x10
-14 J λ = hc/E = (6.626x10
-34 J)(3x10 8 m/s)/(8.2x10
-14 J)
⇒
λ = 2.42x10
-12 m