Calculating Half-Life and Radioactive Decay

[kbutto]

can help me with this question :S?
If 70% of a radioactive substance remains after one year, find its half-life.

and

Strontium-90 is a radioactive isotope with a half-life of 29 years. If you begin with a sample of 800 units, how long will it take for the amount of radioactivity of the strontium sample to be reduced to
(a) 400 units
(b) 200 units
(c) 1 unit


*it has to do wit exponential growth and decay and maybe... with differential equations

 

[CompuChip]

Welcome to PF kbutto.
The half-life t_1/2 is the time until only 50% remains. After two half-lifes, 25% remains, after three, 12,5%, etc.
The general formula we can infer is then: after n half life periods the sample has been reduced by a factor 2ⁿ, so the percentage that is left is then 1/2ⁿ (after 0 half lifes, this gives 1 = 100%, after 1 half-life it gives 1/2, after 2 it gives 1/4, etc.)

Using this, you can solve both questions.

a) After how many half-lives is there 70% left?
b) How many half-lives have passed once you have gone from 800 to 400 units?

 

[HallsofIvy]

can help me with this question :S?
If 70% of a radioactive substance remains after one year, find its half-life.

70%= 0.7= (1/2)^x. What is x?

and

Strontium-90 is a radioactive isotope with a half-life of 29 years. If you begin with a sample of 800 units, how long will it take for the amount of radioactivity of the strontium sample to be reduced to
(a) 400 units
(b) 200 units
(c) 1 unit

400= (1/2)^x*800. 200= (1/2)^y*800. 1= (1/2)^[z*800. What are x, y, and z?

*it has to do wit exponential growth and decay and maybe... with differential equations

Yes, so you may need to use logarithms.

 

[kbutto]

70%= 0.7= (1/2)^x. What is x? 400= (1/2)^x*800. 200= (1/2)^y*800. 1= (1/2)^[z*800. What are x, y, and z?
Yes, so you may need to use logarithms.

I worked it out this way, is it correct?

a) If only 70% of a substance remains, then the final amount y = 70%yo, and it's given that time is one year.

y = 70%yo = yoe^(-k(1))

0.70yo = yoe^(-k)

0.70 = e^(-k)

k = -ln(0.70) = 0.3567 ; Half-life constant of the substance.

Now use this to find the half-life, the final amount should be half original:

y = (1/2)yo = yoe^(-0.3567t)

(1/2)yo = yoe^(-0.3567t)

0.5 = e^(-0.3567t)

t = ln(0.5) / -0.3567 ≈ 1.9 ≈ 2 years

b) For strontium, use the values given to first find the half-life constant of the strontium.

y = (1/2)yo = yoe^(-29t)

0.5 = e^(-29t)

t = ln(0.5) / -29 ≈ 0.024 ; Half-life constant of strontium

now use this to find the time for each case:

1) y = yoe^(-0.024t)

400 = 800e^(-0.024t)

t = ln(0.5) / -0.024 ≈ 28.88 ≈ 29 years

2) y = yoe^(-0.024t)

200 = 800e^(-0.024t)

t = ln(0.25) / -0.024 ≈ 57.76 ≈ 58 years

3) y = yoe^(-0.024t)

1 = 800e^(-0.024t)

t = ln(1/800) / -0.024 ≈ 278.52 ≈ 279 years

yo= y subscript0

 

[HallsofIvy]

I worked it out this way, is it correct?

a) If only 70% of a substance remains, then the final amount y = 70%yo, and it's given that time is one year.

y = 70%yo = yoe^(-k(1))

0.70yo = yoe^(-k)

0.70 = e^(-k)

k = -ln(0.70) = 0.3567 ; Half-life constant of the substance.

Now use this to find the half-life, the final amount should be half original:

y = (1/2)yo = yoe^(-0.3567t)

(1/2)yo = yoe^(-0.3567t)

0.5 = e^(-0.3567t)

t = ln(0.5) / -0.3567 ≈ 1.9 ≈ 2 years

Yes, that's correct but isn't this simpler? Saying the half life is T means the quantity is multiplied by 1/2 every half-life. There are t/T "half-lives" in time t. In one year, t= 1 so 0.7= (0.5)^1/T so ln(0.7)= ln(0.5^1/T)= ln(0.5)/T. T= ln(0.5)/ln(0.7)= 1.94

b) For strontium, use the values given to first find the half-life constant of the strontium.

y = (1/2)yo = yoe^(-29t)

0.5 = e^(-29t)

t = ln(0.5) / -29 ≈ 0.024 ; Half-life constant of strontium

now use this to find the time for each case:

1) y = yoe^(-0.024t)

400 = 800e^(-0.024t)

t = ln(0.5) / -0.024 ≈ 28.88 ≈ 29 years

2) y = yoe^(-0.024t)

200 = 800e^(-0.024t)

t = ln(0.25) / -0.024 ≈ 57.76 ≈ 58 years

3) y = yoe^(-0.024t)

1 = 800e^(-0.024t)

t = ln(1/800) / -0.024 ≈ 278.52 ≈ 279 years

yo= y subscript0

But WHY use exponentials for this simple problem?
400= 1/2 (800) so this requires exactly one half-life. Which you were told is 29 years.

200 is 1/2(400) so this is another half-life: 29+ 29= 58 years.
Or 200= (1/4)(800)= (1/2)²(800) so 2 half-lives: 58 years.

The last is a little harder: 1= 800(1/2)^x so (1/2)^x= 1/800, which is not an integer power of 2. x ln(1/2)= ln(1/800) x= ln(1/800)/ln(1/2)= ln(800)/ln(2)= 9.64 half-lives or 9.64(29)= 280 years.