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The Nucleus: A Chemist’s ViewChapter 20E-mail: [email protected]
Web-site: http://clas.sa.ucsb.edu/staff/terri/
The Nucleus: A Chemist’s View1. 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
The Nucleus: A Chemist’s View2. 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 View3. 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?
What is the daughter nucleus if 242U underwent a decay series producing 4 alpha and 3 beta particles?
The Nucleus: A Chemist’s View4. 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 View5. 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 View6. 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 View7. 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 View8. 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 Na3
32PO4 after 35.0 days? Assume the atomic mass of 32P is 32.0.
The Nucleus: A Chemist’s View9. 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 View10. 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 View11. 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 View1. Supply the missing particle for each of the following nuclear
reactions:
a. 73Ga 73Ge + ? ⇒ beta particle
b. 192 Pt 188Os + ? ⇒ alpha particle
c. 205Bi 205Pb + ? ⇒ positron
d. 241Cm + ? 241Am ⇒ electron
The Nucleus: A Chemist’s View2. Write an equation for each of the following:
a. 68Ga undergoing electron capture 68Ga + 68Zn
b. 62Cu undergoing positron emission 62Cu + 62Ni
c. 212Fr undergoing alpha decay 212Fr + 208At
d. 129Sb undergoing beta decay129Sb + 52Te
The Nucleus: A Chemist’s View3. 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 + Y
247 = 207 + 4X
X = 10
97 = 82 + 2(10) –Y
Y = 5
What is the daughter nucleus if 242U underwent a decay series producing 4 alpha and 3 beta particles?
242U 4 + 3 + 226Fr
The Nucleus: A Chemist’s View4. 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 View5. 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 =
= or =
= = 0.2 or 20% remains
The Nucleus: A Chemist’s View6. 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 a strong force in the nucleus
ΔE = Δmc2
The Nucleus: A Chemist’s View7. 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
k = ⇒ t1/2 = = = 64.4hr
The Nucleus: A Chemist’s View8. 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 Na3
32PO4 after 35.0 days? Assume the atomic mass of 32P is 32.0.
In the original sample ⇒ 175 mg Na332PO4 = 33.95 mg 32P
using ln[A] = -kt + ln[A]o and t1/2 = 0.693/k
ln[A] = + ln[A]o
ln[A] = + ln(33.95 mg)
[A] = 6.22 mg
The Nucleus: A Chemist’s View9. 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
t = = = 3.4x109 yr
The Nucleus: A Chemist’s View10. 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 +
ΔE = Δmc2
ΔE = [26(1.67262 x 10-27 kg) + 30(1.67493 x 10-27 kg) –(55.9349 amu - 26(9.10939 x 10-31 kg)](3x108m/s)2
ΔE = = 1.407x10-12 J/nucleon
The Nucleus: A Chemist’s View11. 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 + 2
Since the mass of an electron and a positron are the same then the energy of the 2 photons are 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