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### Topic: Half-life  (Read 14445 times)

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#### khalid

• Guest ##### Half-life
« on: July 27, 2005, 08:38:31 PM »
Hello, I'm new to the forum, nice to meet you all ^^

My question:

the Isotop Sr90 has a half life of 28.1 years. how much would remain from a 1g of Sr after 25 years.

I toke the topic of half-life in geology and general chem before, but i dont have my textbook at the moment to review and find the answer ><

By dividing 28.1/25, this would give us the number of half lifes in 28.1 years i think?

But i dont know how to take it from there, and how to find the mass after 28.1 years.

A 2nd part of the question asks about the percentage of the initial acivities remains after 25 years. i found the decay constant, k in 1/sec., but couldnt continou.

I would ask the instractor if i could, but am taking this course as an independent study. sorry if my question was not very clear, english is my 2nd language.

Thanks alot and i hope to hear from you soon.
« Last Edit: July 27, 2005, 08:40:14 PM by khalid »

#### sdekivit

• Chemist
• Full Member
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• Gender: • B.Sc Biomedical Sciences, Utrecht University ##### Re:Half-life
« Reply #1 on: July 28, 2005, 03:44:49 AM »
Hello, I'm new to the forum, nice to meet you all ^^

My question:

the Isotop Sr90 has a half life of 28.1 years. how much would remain from a 1g of Sr after 25 years.

I toke the topic of half-life in geology and general chem before, but i dont have my textbook at the moment to review and find the answer ><

By dividing 28.1/25, this would give us the number of half lifes in 28.1 years i think?

But i dont know how to take it from there, and how to find the mass after 28.1 years.

A 2nd part of the question asks about the percentage of the initial acivities remains after 25 years. i found the decay constant, k in 1/sec., but couldnt continou.

I would ask the instractor if i could, but am taking this course as an independent study. sorry if my question was not very clear, english is my 2nd language.

Thanks alot and i hope to hear from you soon.

the formula for radioactive decay is:

N(t) = N(0) * (1/2)^(t/tau) with N = numbers of cores and tau = half life time.

so 1g Sr = 0,0114129194 mol = 6,8706 x 10^21 cores.

so: N(25 years) = 6,8706 x 10^21 * (1/2)^25/28,1
--> N(25 years) = 3,7083 x 10^21 cores

--> that's 0,0061599 mol Sr = 0,540 g Sr.

Same for the activities of a radioactive isotope:

A(t) = A(0) * (1/2)^(t/tau)

we want the percentage of activity f the initial activities after 25 years, so that A(25)/A(0) * 100%

so the percentage is given in: (1/2)^(t/tau)

--> 1/2^(25/28,1) * 100% = 54 %

#### Mitch ##### Re:Half-life
« Reply #2 on: July 28, 2005, 03:50:38 AM »
The radioactive decay formula is actually.

N(t) = N(0) * e-0.693t/tau If you define tau as the half life.
« Last Edit: July 28, 2005, 03:51:08 AM by Mitch »
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#### sdekivit

• Chemist
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• Gender: • B.Sc Biomedical Sciences, Utrecht University ##### Re:Half-life
« Reply #3 on: July 28, 2005, 04:07:20 AM »
The radioactive decay formula is actually.

N(t) = N(0) * e-0.693t/tau If you define tau as the half life.

it's the same as you have it in ln-form and i have it simple in log-form.

here's a derivation

ln x = 2,303 log x

so we have N(t) = N(0) * (1/2)^(t/tau)

take the log on both sides:

log N(t) = log N(0) + (t/tau)log 0,5

convert to ln:

2,303 log N(t) = 2,303 log N(0) + 2,303(t/tau)log 0,5

thus:

ln N(t) = ln N(0) + (t/tau) ln 0,5 --> ln 0,5 = -0,693

now both sides e^:

N(t) = N(0) * e^-0,693(t/tau)
« Last Edit: July 28, 2005, 04:33:24 AM by sdekivit »

#### Mitch ##### Re:Half-life
« Reply #4 on: July 28, 2005, 12:22:29 PM »
Best to keep things in the natural log form. It flows nicely from the differential form of the decay law.
Most Common Suggestions I Make on the Forums.
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2. Don't confuse thermodynamic stability with chemical reactivity.
3. Forum Supports LaTex

#### sdekivit

• Chemist
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• Mole Snacks: +32/-3
• Gender: • B.Sc Biomedical Sciences, Utrecht University ##### Re:Half-life
« Reply #5 on: July 28, 2005, 12:30:44 PM »
Best to keep things in the natural log form. It flows nicely from the differential form of the decay law.

i agree, that form the differential equation dN/dt = -lambda * N we'll get a power of e, but when you don't know how to solve differential equations, it's easier to say that the decay is exponential with a constant factor: 1/2

--> from maths, the formula for a exponential decay ( dutch variables used)

H = b * g^t --> N = N(0) x (1/2)^tau

to let it count for every t, just remember that 3 months for example is 3/12 year and thus we get:

N = N(0) x (1/2)^(t/tau)

it's not wrong and easier for people that don't know differential equations. But again, i agree that solving the differential equation is better.

(here in holland we get radioactive decay in secondary school without ever seeing any differential equation. That's why i did it this way. The answer will still be the same )
« Last Edit: July 28, 2005, 12:32:17 PM by sdekivit »

#### Mitch ##### Re:Half-life
« Reply #6 on: July 28, 2005, 05:35:56 PM »
Whatever floats your boat, just don't get the youth confused.
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
3. Forum Supports LaTex

#### khalid

• Guest ##### Re:Half-life
« Reply #7 on: July 29, 2005, 04:54:27 PM »
Thanks alot for the help.

I was able to solve all 7 problems with the info you gave me. ^^;