So I'm studying for a chemistry final and I don't understand this question.
I need help because a large portion of the test is about the half-life of atoms.
Here's the question:
Element A Decays into element B. The half-life of element A is 3 years. You have a sample that contains 50 atoms of A and 3150 of B.
A) how many atoms of element A were originally present?
B) how many years has it been since the samples started to decay?
Thank you so much for helping
Chemistry help?
2016-05-10 6:55 am
回答 (2)
2016-05-10 7:11 am
A)
Initially, there are only atoms of element A and no atom of element B.
1 atom of element A decays to form 1 atom of element B.
Number of atoms of element A originally present
= 50 + 3150
= 3200
B)
Let n be the number of half-lives since the sample started to decay.
Original number of atoms of element A = 3200
Now number of atoms of element A = 50
Fraction of atoms of element A left unchanged :
(1/2)ⁿ = 50/3200
(1/2)ⁿ = (1/64)
(1/2)ⁿ = (1/2)⁶
n = 6
Time taken since the sample started to decay = (3 years) × 6 = 18 years
Initially, there are only atoms of element A and no atom of element B.
1 atom of element A decays to form 1 atom of element B.
Number of atoms of element A originally present
= 50 + 3150
= 3200
B)
Let n be the number of half-lives since the sample started to decay.
Original number of atoms of element A = 3200
Now number of atoms of element A = 50
Fraction of atoms of element A left unchanged :
(1/2)ⁿ = 50/3200
(1/2)ⁿ = (1/64)
(1/2)ⁿ = (1/2)⁶
n = 6
Time taken since the sample started to decay = (3 years) × 6 = 18 years
2016-05-10 7:06 am
Use the law of activity A(n)=A°/2^n
where A(n) is the activity after n periods (or half lives) have occured, A° the initial activity.
Activity can be simplified and you can use the quantity of matter as a pretty good approximation.
So, now that is said, let's look.
You have 50 A and 3150 B. If initially, you had only A (not stated in your problem, but let's assume that's the case), then all the Bs come from decomposition of A, so you had initially 3200 A.
That gives you A(n) = 50 and A°=3200
2^n = A°/A(n)
n.ln(2) = ln(A°/A(n))
n = ln(A°/A(n)) / ln(2)
n = ln(3200/50)/ln(2)
n = 6
so 6 half lives have passed by, that makes 18 years.
where A(n) is the activity after n periods (or half lives) have occured, A° the initial activity.
Activity can be simplified and you can use the quantity of matter as a pretty good approximation.
So, now that is said, let's look.
You have 50 A and 3150 B. If initially, you had only A (not stated in your problem, but let's assume that's the case), then all the Bs come from decomposition of A, so you had initially 3200 A.
That gives you A(n) = 50 and A°=3200
2^n = A°/A(n)
n.ln(2) = ln(A°/A(n))
n = ln(A°/A(n)) / ln(2)
n = ln(3200/50)/ln(2)
n = 6
so 6 half lives have passed by, that makes 18 years.
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