C. Johannesson CHAPTER 22 Nuclear Chemistry II. Radioactive Decay (p. 705 - 712) II. Radioactive...
-
Upload
alexandrina-carpenter -
Category
Documents
-
view
214 -
download
0
Transcript of C. Johannesson CHAPTER 22 Nuclear Chemistry II. Radioactive Decay (p. 705 - 712) II. Radioactive...
C. Johannesson
CHAPTER 22
Nuclear
Chemistry
CHAPTER 22
Nuclear
ChemistryII. Radioactive II. Radioactive
DecayDecay(p. 705 - 712)
II. Radioactive II. Radioactive DecayDecay
(p. 705 - 712)
I
IV
III
II
C. Johannesson
He42
A. Types of RadiationA. Types of RadiationA. Types of RadiationA. Types of Radiation
Alpha particle () helium nucleus paper2+
Beta particle (-) electron e0
-11-
leadPositron (+)
positron e01
1+
Gamma () high-energy photon 0
concrete
C. Johannesson
B. Nuclear DecayB. Nuclear DecayB. Nuclear DecayB. Nuclear Decay
Alpha Emission
He Th U 42
23490
23892
parentnuclide
daughternuclide
alphaparticle
Numbers must balance!!
C. Johannesson
B. Nuclear DecayB. Nuclear DecayB. Nuclear DecayB. Nuclear Decay
Beta Emission
e Xe I 0-1
13154
13153
electronPositron Emission
e Ar K 01
3818
3819
positron
C. Johannesson
B. Nuclear DecayB. Nuclear DecayB. Nuclear DecayB. Nuclear Decay
Electron Capture
Pd e Ag 10646
0-1
10647
electronGamma Emission
Usually follows other types of decay.
Transmutation One element becomes another.
C. Johannesson
B. Nuclear DecayB. Nuclear DecayB. Nuclear DecayB. Nuclear Decay
Why nuclides decay… need stable ratio of neutrons to protons
He Th U 42
23490
23892
e Xe I 0-1
13154
13153
e Ar K 01
3818
3819
Pd e Ag 10646
0-1
10647
DECAY SERIES TRANSPARENCY
C. Johannesson
C. Half-lifeC. Half-lifeC. Half-lifeC. Half-life
Half-life (t½) Time required for half the atoms of a
radioactive nuclide to decay. Shorter half-life = less stable.
C. Johannesson
C. Half-lifeC. Half-lifeC. Half-lifeC. Half-life
nif mm )( 2
1
mf: final massmi: initial massn: # of half-lives
C. Johannesson
C. Half-lifeC. Half-lifeC. Half-lifeC. Half-life Fluorine-21 has a half-life of 5.0 seconds. If you start
with 25 g of fluorine-21, how many grams would remain after 60.0 s?
GIVEN:
t½ = 5.0 s
mi = 25 g
mf = ?
total time = 60.0 s
n = 60.0s ÷ 5.0s =12
WORK:
mf = mi (½)n
mf = (25 g)(0.5)12
mf = 0.0061 g