more radioactive decay calculations
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More Radioactive Decay Calculations. Problem #1. The half-life of cobalt-60 is 5.3 years. How much of a 1.000 mg sample of cobalt-60 is left after a 15.9 yr period?. Using A = A o (1/2) n. A = final amount A o = initial amount n = number of half-lives elapsed. - PowerPoint PPT PresentationTRANSCRIPT
More Radioactive Decay Calculations
More Radioactive Decay Calculations
Problem #1Problem #1
The half-life of cobalt-60 is 5.3 years. How much of a 1.000 mg sample of cobalt-60 is left after a 15.9 yr period?
The half-life of cobalt-60 is 5.3 years. How much of a 1.000 mg sample of cobalt-60 is left after a 15.9 yr period?
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = final amountAo = initial amountn = number of half-lives elapsed
A = final amountAo = initial amountn = number of half-lives elapsed
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = final amountAo = initial amountn = number of half-lives elapsed
Calculate n:n = T/t1/2
T = total time elapsedt 1/2 = half-life of atom
A = final amountAo = initial amountn = number of half-lives elapsed
Calculate n:n = T/t1/2
T = total time elapsedt 1/2 = half-life of atom
A = Ao(1/2)nA = Ao(1/2)n
A = ?Ao = 1.000 mgn = ?
A = ?Ao = 1.000 mgn = ?
A = Ao(1/2)nA = Ao(1/2)n
A = ?Ao = 1.000 mgn = ?
T = 15.9 yrst 1/2 = 5.3 yrs
n = T/t1/2 = 15.9/5.3 = 3
A = ?Ao = 1.000 mgn = ?
T = 15.9 yrst 1/2 = 5.3 yrs
n = T/t1/2 = 15.9/5.3 = 3
A = Ao(1/2)nA = Ao(1/2)n
A = ?Ao = 1.000 mgn = 3
n = T/t1/2 = 15.9/5.3 = 3
A = ?Ao = 1.000 mgn = 3
n = T/t1/2 = 15.9/5.3 = 3
A = Ao(1/2)nA = Ao(1/2)n
A = ?Ao = 1.000 mgn = 3
X = 1.000 mg (.5)3 Make certain that you try this with your
calculator!!!
A = ?Ao = 1.000 mgn = 3
X = 1.000 mg (.5)3 Make certain that you try this with your
calculator!!!
A = Ao(1/2)nA = Ao(1/2)n
A = ?Ao = 1.000 mgn = 3
X = 1.000 mg (.5)3
X = .125 mg
A = ?Ao = 1.000 mgn = 3
X = 1.000 mg (.5)3
X = .125 mg
Rinky think method, same problem…Rinky think method, same problem…
total time elapsed[---------------15.9 years------------]
total time elapsed[---------------15.9 years------------]
Rinky think method…Rinky think method… total time elapsed[---------------15.9 years------------]
How many half-lives fit in this time frame?
[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
total time elapsed[---------------15.9 years------------]
How many half-lives fit in this time frame?
[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
Rinky think method…Rinky think method… total time elapsed[---------------15.9 years------------]
How many half-lives fit in this time frame?[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
15.9 yrs/5.3 yrs = 3 half-lives
total time elapsed[---------------15.9 years------------]
How many half-lives fit in this time frame?[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
15.9 yrs/5.3 yrs = 3 half-lives
Rinky think method…Rinky think method… total time elapsed[---------------15.9 years------------]
How many half-lives fit in this time frame?[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
Decrease mass by half for each half-life.
total time elapsed[---------------15.9 years------------]
How many half-lives fit in this time frame?[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
Decrease mass by half for each half-life.
Rinky think method…Rinky think method… total time elapsed[---------------15.9 years------------]
How many half-lives fit in this time frame?[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
Decrease mass by half for each half-life.
1 mg .5 mg .25mg .125 mg!
total time elapsed[---------------15.9 years------------]
How many half-lives fit in this time frame?[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
Decrease mass by half for each half-life.
1 mg .5 mg .25mg .125 mg!
Rinky think method…Rinky think method… total time elapsed[---------------15.9 years------------]
[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
1 mg .5 mg .25mg .125 mg!
total time elapsed[---------------15.9 years------------]
[--5.3 yrs--][--5.3 yrs--][--5.3 yrs--]
1 mg .5 mg .25mg .125 mg!
Problem #2Problem #2
• If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 years. What is the half-life of strontium-90?
• If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 years. What is the half-life of strontium-90?
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = .95 gAo = 1.000 gn = ?
Calculate n:n = T/t1/2
T = 2 yearst 1/2 = ?
A = .95 gAo = 1.000 gn = ?
Calculate n:n = T/t1/2
T = 2 yearst 1/2 = ?
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = .95 gAo = 1.000 gn = 2 yrs/x
Calculate n:n = T/t1/2
T = 2 yearst 1/2 = ?
A = .95 gAo = 1.000 gn = 2 yrs/x
Calculate n:n = T/t1/2
T = 2 yearst 1/2 = ?
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = .95 gAo = 1.000 gn = 2 yrs/x
Plug in the information:
.95 g = 1 g (.5)2 yrs/x
A = .95 gAo = 1.000 gn = 2 yrs/x
Plug in the information:
.95 g = 1 g (.5)2 yrs/x
Using A = Ao(1/2)nUsing A = Ao(1/2)n
.95 g = 1 g (.5)2 yrs/x
Simplify by dividing both sides by 1 g.
.95 = .52 yrs/x
.95 g = 1 g (.5)2 yrs/x
Simplify by dividing both sides by 1 g.
.95 = .52 yrs/x
Using A = Ao(1/2)nUsing A = Ao(1/2)n
.95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
To get the exponent in a solvable position, take the logarithm of the problem:
log.95 = 2 yrs/x (log.5)
.95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
To get the exponent in a solvable position, take the logarithm of the problem:
log.95 = 2 yrs/x (log.5)
Using A = Ao(1/2)nUsing A = Ao(1/2)n
95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
log.95 = 2 yrs/x (log.5)
Simplify by dividing both sides by log.5
log.95/log.5 = 2 yrs/x
95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
log.95 = 2 yrs/x (log.5)
Simplify by dividing both sides by log.5
log.95/log.5 = 2 yrs/x
Using A = Ao(1/2)nUsing A = Ao(1/2)n
.95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
log.95 = 2 yrs/x (log.5)log.95/log.5 = 2 yrs/x
Calculate the logs and divide them:
X(.074) = 2 yrs
.95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
log.95 = 2 yrs/x (log.5)log.95/log.5 = 2 yrs/x
Calculate the logs and divide them:
X(.074) = 2 yrs
Using A = Ao(1/2)nUsing A = Ao(1/2)n
.95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
log.95 = 2 yrs/x log.5log.95/log.5 = 2 yrs/xX(.074) = 2 yrs
Divide by 0.74 to solve for x:
X = 2 yrs/.074
.95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
log.95 = 2 yrs/x log.5log.95/log.5 = 2 yrs/xX(.074) = 2 yrs
Divide by 0.74 to solve for x:
X = 2 yrs/.074
Using A = Ao(1/2)nUsing A = Ao(1/2)n
.95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
log.95 = 2 yrs/x log.5log.95/log.5 = 2 yrs/xX(.074) = 2 yrsX = 2 yrs/.074X = 27 years
.95 g = 1 g (.5)2 yrs/x
.95 = .52 yrs/x
log.95 = 2 yrs/x log.5log.95/log.5 = 2 yrs/xX(.074) = 2 yrsX = 2 yrs/.074X = 27 years
Rinky think methodRinky think method
Quantities too small for this problem!
All problems on the test will be okay for rinky thinking…
Quantities too small for this problem!
All problems on the test will be okay for rinky thinking…
Problem #3Problem #3A wooden object from an archeological siteis subjected to radiocarbon dating. The activityof the sample due to carbon-14 is measured to be.43 disintegrations per second. The activity of acarbon sample of equal mass from fresh wood is15.2 disintegrations per second. The half-lifeof carbon-14 is 5715 yr. What is the age ofthe archeological sample?
A wooden object from an archeological siteis subjected to radiocarbon dating. The activityof the sample due to carbon-14 is measured to be.43 disintegrations per second. The activity of acarbon sample of equal mass from fresh wood is15.2 disintegrations per second. The half-lifeof carbon-14 is 5715 yr. What is the age ofthe archeological sample?
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
Calculate n:n = T/t1/2
T = ?t 1/2 = 5715 years
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
Calculate n:n = T/t1/2
T = ?t 1/2 = 5715 years
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.0283 = (.5) x/5715 yrs
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.0283 = (.5) x/5715 yrs
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.028 = (.5) x/5715 yrs
log.028 = x/5715 yrs( log.5)
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.028 = (.5) x/5715 yrs
log.028 = x/5715 yrs( log.5)
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.028 = (.5) x/5715 yrs
log.028/ log.5 = x/5715 yrs
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.028 = (.5) x/5715 yrs
log.028/ log.5 = x/5715 yrs
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.028 = (.5) x/5715 yrs
log.028/ log.5 = x/5715 yrs5.16 = x/5715 yrs
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.028 = (.5) x/5715 yrs
log.028/ log.5 = x/5715 yrs5.16 = x/5715 yrs
Using A = Ao(1/2)nUsing A = Ao(1/2)n
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.028 = (.5) x/5715 yrs
log.028/ log.5 = x/5715 yrs5.16 = x/5715 yrsX = 29395.6 yrs
A = .43 disintegrationsAo = 15.2 disintegrationsn = x/5715 yrs
.43 disintegrations = 15.2 disintegrations(.5) x/5715 yrs
.028 = (.5) x/5715 yrs
log.028/ log.5 = x/5715 yrs5.16 = x/5715 yrsX = 29395.6 yrs
Rinky think method…Rinky think method…15.2 disintegrations when start so
divide by 2 until reach the final amount of .43 (or close to it!).
15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475
15.2 disintegrations when start so divide by 2 until reach the final amount of .43 (or close to it!).
15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475
Rinky think method…Rinky think method…15.2 disintegrations when start so
divide by 2 until reach the final amount of .43 (or close to it!).
15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475
Count how many half-lives have elapsed…
15.2 disintegrations when start so divide by 2 until reach the final amount of .43 (or close to it!).
15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475
Count how many half-lives have elapsed…
Rinky think method…Rinky think method…15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --
> .475
1 2 3 4 5+
Count how many half-lives have elapsed…
15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475
1 2 3 4 5+
Count how many half-lives have elapsed…
Rinky think method…Rinky think method…15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --
> .475
1 2 3 4 5+
Multiply half-life of carbon-14 by the number of half-lived elapsed.
15.2 --> 7.6 --> 3.8 --> 1.9 --< .95 --> .475
1 2 3 4 5+
Multiply half-life of carbon-14 by the number of half-lived elapsed.
Rinky think method…Rinky think method…15.2 --> 7.6 --> 3.8 --> 1.9 --> .95 --> .475
1 2 3 4 5+ a little
5715 years x 5 = 28,575 years + a little
This answer is somewhat close to the formula method, thus an acceptable answer on the test!
15.2 --> 7.6 --> 3.8 --> 1.9 --> .95 --> .475
1 2 3 4 5+ a little
5715 years x 5 = 28,575 years + a little
This answer is somewhat close to the formula method, thus an acceptable answer on the test!