The Nucleus: A Chemist’s ViewChapter 20E-mail: benzene4president@gmail.comWeb-site: http://clas.sa.ucsb.edu/staff/terri/

The Nucleus: A Chemist’s View1. Write an equation for each of the following:

a. 68Ga undergoing electron captureb. 62Cu undergoing positron emissionc. 212Fr undergoing alpha decayd. 129Sb undergoing beta decay

The Nucleus: A Chemist’s ViewType of Decay

Nuclear Problem

Nuclear Transformation

Emitted Particle

Danger Level Example

Alpha (α) Nucleus is too big (Z > 84)

Lose 2 and 2 A–4 and Z–2 Low +

Beta (β–) Neutron rich or : is too high

Convert neutron into proton and electron

+ Z+1

or β– Medium +

Positron Emission

(β+)

Proton rich or : is too high

Convert proton into neutron and positron

+ Z–1

or β+ High +

Electron Capture

(EC)

Proton rich or : is too high

Convert proton into neutron by absorbing

an electron

Z–1

None Low

Gamma Emission(γ)

Nucleus has too much energy

Lose energy in the form of gamma

photons

or gamma photon High * +

The Nucleus: A Chemist’s View2. 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 View3. 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 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. 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 View7. 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 View8. 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 View9. 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 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. 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 View2. Supply the missing particle for each of the following nuclear

reactions:a. 73Ga 73Ge + ? ⇒ beta particleb. 192 Pt 188Os + ? ⇒ alpha particlec. 205Bi 205Pb + ? ⇒ positrond. 241Cm + ? 241Am ⇒ electron

The Nucleus: A Chemist’s View3. What is the daughter nucleus if 242U underwent a decay series

producing 4 alpha and 3 beta particles? 242U 4 + 3 + 226Fr

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 + Y247 = 207 + 4XX = 1097 = 82 + 2(10) –Y Y = 5

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. 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 View7. 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 Na3

32PO4 = 33.95 mg 32Pusing ln[A] = -kt + ln[A]o and t1/2 = 0.693/kln[A] = + ln[A]o

ln[A] = + ln(33.95 mg)[A] = 6.22 mg

The Nucleus: A Chemist’s View8. 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/tt = = = 3.4x109 yr

The Nucleus: A Chemist’s View9. 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 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 an electron and a positron have the same mass then the energy of the 2 photons is also the sameΔE = Δmc2 ΔE = (0 kg - 9.10939x10-31kg)(3x108 m/s)2

ΔE = - 8.2x10-14Jλ = hc/E = (6.626x10-34J)(3x108 m/s)/(8.2x10-14J) ⇒ λ = 2.42x10-12m