Half-Life NotesHalf-Life Notes

Headsium DecayHeadsium Decay# of

½ livesIndividual Lab

IslandClass Course

0

1

2

3

4

3030

½ Life Activity ½ Life Activity (Bring back Calculators)(Bring back Calculators)

1.1. Go back to the lab with your partner Go back to the lab with your partner and grab a ½ Life activity kit for and grab a ½ Life activity kit for EACH of youEACH of you

2.2. You all have 30 pennies (radioactive You all have 30 pennies (radioactive isotopes) to begin with. Place them isotopes) to begin with. Place them all HEADS UP. Shake the box (each all HEADS UP. Shake the box (each trial = ½ life).trial = ½ life).

3.3. Take out the tails (these are the Take out the tails (these are the decayed isotopes). Record how decayed isotopes). Record how many “headsium” isotopes remainmany “headsium” isotopes remain

4.4. Repeat 3 more times and record Repeat 3 more times and record resultsresults

The Half-LifeThe Half-LifeThe The half-life Thalf-life T1/21/2

of an isotope is of an isotope is the time in the time in which one-half which one-half of its unstable of its unstable nuclei will nuclei will decay.decay.

No

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2

N

0

4

N

Number of Half-lives

Nu

mb

er

Un

deca

yed

N

ucl

ei

1 432

0

1

2

n

N N

0

1

2

n

N N

Where Where n n is is number of half-number of half-

liveslives Number of Half-Number of Half-lives:lives:

12

tnT

1

2

tnT

Example 1:Example 1: A sample of A sample of iodine-131iodine-131 has has a half life of a half life of 88 daysdays. If you start out with . If you start out with 15.0 grams of I-131, how much is left of 15.0 grams of I-131, how much is left of the sample the sample 3232 daysdays later? later?

First we determine the number of half-First we determine the number of half-lives:lives:

12

tnT

1

2

tnT

n = 32 days/8days = 4 half-n = 32 days/8days = 4 half-lives lives

0

1

2

n

N N

0

1

2

n

N N

N = (15.0g) (1/2)N = (15.0g) (1/2)44 = 0.938 grams = 0.938 grams remaining remaining

15.0 g 7.50 3.75 1.88 0.938 15.0 g 7.50 3.75 1.88 0.938 g g

Example 2:Example 2: Cobalt-60Cobalt-60 is a radioisotope is a radioisotope used as a source of ionizing radiation in used as a source of ionizing radiation in cancer treatment; the radiation it cancer treatment; the radiation it emits is effective in killing rapidly emits is effective in killing rapidly dividing cancer cells. dividing cancer cells. Co-60 has a ½ Co-60 has a ½ life of five yearslife of five years..

a.a.If the hospital starts with a If the hospital starts with a 1000. mg 1000. mg supply, how supply, how many mg of Co-60 would they have to purchase after many mg of Co-60 would they have to purchase after 10 yrs10 yrs. to replenish their original supply?. to replenish their original supply?

b.b.How many ½ lives would it take for the supply to Co-60 to dwindle to How many ½ lives would it take for the supply to Co-60 to dwindle to 1.1.Less than 10.0 % of the original?Less than 10.0 % of the original?2.2.Less ant 1.00 % of the original?Less ant 1.00 % of the original?3.3.Less than .100 % of the original?Less than .100 % of the original?

a. If the hospital starts with a 1000. mg supply, how a. If the hospital starts with a 1000. mg supply, how many mg of Co-60 would they have to purchase after many mg of Co-60 would they have to purchase after 10 yrs. to replenish their original supply?10 yrs. to replenish their original supply?

First we determine the number of half-First we determine the number of half-lives:lives:

NN00 = 1000. mg = 1000. mg12

tnT

1

2

tnT

= 10 yrs/5 yrs = 2 half-lives = 10 yrs/5 yrs = 2 half-lives

0

1

2

n

N N

0

1

2

n

N N

=(1000. mg) (1/2)=(1000. mg) (1/2)22 = 250.0 mg = 250.0 mg remainremain

1000. mg – 250.0 mg = 750. mg needed to 1000. mg – 250.0 mg = 750. mg needed to replenishreplenish

1000. -> 500.0 -> 250.0 mg 1000. -> 500.0 -> 250.0 mg remainremain

b.b.How many ½ lives would it take for the supply to Co-60 to dwindle to How many ½ lives would it take for the supply to Co-60 to dwindle to 1.1.Less than 10.0 % of the original?Less than 10.0 % of the original?2.2.Less ant 1.00 % of the original?Less ant 1.00 % of the original?3.3.Less than .100 % of the original?Less than .100 % of the original?

1. “1. “44” - ” - 50% 50% left after 1 half-life, left after 1 half-life, 25 %25 % left after 2 half-lives, left after 2 half-lives, 12.5 % 12.5 % left after 3 left after 3 half-lives, and half-lives, and 6.256.25 % after 4 half-lives % after 4 half-lives

2. “2. “77” - ” - .781 .781 % left after 7 half-lives% left after 7 half-lives

3. “3. “1010” - ” - .0977 .0977 % left after 10 half-% left after 10 half-liveslives

Section 24.2Section 24.2

16. 16. CalculateCalculate how much of a how much of a 10.0 g10.0 g sample of sample of americium-241 remains americium-241 remains after four half-livesafter four half-lives. .

Americium-241is a radioisotope commonly used in Americium-241is a radioisotope commonly used in smoke detectors and has a half-life of 430 ysmoke detectors and has a half-life of 430 y

0.625 g0.625 g10.0→5.00→2.50→1.25→10.0→5.00→2.50→1.25→

0

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2

n

N N

0

1

2

n

N N

=(10.0 g) (1/2)=(10.0 g) (1/2)44 = 0.625 mg = 0.625 mg remainremain

17. After 17. After 2.00 y, 1.986 g 2.00 y, 1.986 g of a radioisotope remains from a sample of a radioisotope remains from a sample that had an original mass of that had an original mass of 2.000 g2.000 g

A. Calculate the half-lifeA. Calculate the half-life N = NN = N00(1/2)(1/2)t/Tt/T

1.986 g = 2.000 g x (1/2)1.986 g = 2.000 g x (1/2)(2.00 y)/T(2.00 y)/T

1.986 g1.986 g = = (1/2)(1/2)(2.00 y)/T(2.00 y)/T 0.9930 = 0.50.9930 = 0.5(2.00 y)/T(2.00 y)/T

2.000 g2.000 g ln 0.9930 = ln (0.5ln 0.9930 = ln (0.5(2.00 y)/T(2.00 y)/T))

ln 0.9930 = ln (0.5) x (2.00 ln 0.9930 = ln (0.5) x (2.00 y)/Ty)/T

T = T = ln (0.5) x (2.00 y)ln (0.5) x (2.00 y) = = 197 years197 years ln 0.9930 ln 0.9930

Section 24.2Section 24.2

17. After 17. After 2.00 y, 1.986 g 2.00 y, 1.986 g of a radioisotope of a radioisotope remains from a sample that had an original remains from a sample that had an original mass of mass of 2.000 g2.000 g

B. How much of the radioisotope remains B. How much of the radioisotope remains after after 10.00 y10.00 y??

N = 2.000 g x (1/2)N = 2.000 g x (1/2)10.00 y/197 y 10.00 y/197 y = = 1.93 g1.93 g

18. A sample of polonium- 214 18. A sample of polonium- 214 originally has a mass of 1.0 g. originally has a mass of 1.0 g. Express the mass remaining as Express the mass remaining as a percent of the original a percent of the original sample after a period of one, sample after a period of one, two, and three half-lives. Graph two, and three half-lives. Graph the percent remaining versus the percent remaining versus the number of half-lives. the number of half-lives.

Approximately how much time Approximately how much time has elapsed when 20% of the has elapsed when 20% of the original sample remains? original sample remains?

Section 24.2Section 24.2No

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Number of Half-livesN

um

ber

Und

eca

yed

N

ucl

ei

1 432

50% 50%

25% 25% 12.5% 12.5%

~ 2.3 half lives~ 2.3 half lives

From Table 24.5 pg 871: From Table 24.5 pg 871: TT1/21/2= 163.7 = 163.7 μμss

163.7 163.7 x x 2.3 2.3 ~~ 380 380 μμss