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# help please

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A certain element has a half life of 1.5 billion years. a. You find a rock containing a mixture of the element and lead. You determine that 65 ​% of the original element​ remains; the other 35 ​% decayed into lead. How old is the​ rock?

b. Analysis of another rock shows that it contains15​% of its original​ element; the other 85​% decayed into lead. How old is the​ rock?

a. The rock is approximately________billion years old.

​(Round to one decimal place as​ needed.)

b. The rock is approximately ________ billion years old.

Dec 6, 2017

### 1+0 Answers

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A certain element has a half-life of 1.5 billion years. a. You find a rock containing a mixture of the element and lead. You determine that 65 % of the original element remains; the other 35 % decayed into lead. How old is the rock?

Let the original age of the rock =A. Then we have:

65% =100% x 2^-(A/1.5), solve for A

0.65 = 2^(-0.666667 A)

0.65 = 13/20 and 2^(-0.666667 A) = 2^(-2 A/3):
13/20 = 2^(-(2 A)/3)

13/20 = 2^(-(2 A)/3) is equivalent to 2^(-(2 A)/3) = 13/20:
2^(-(2 A)/3) = 13/20

Take reciprocals of both sides:
2^((2 A)/3) = 20/13

Take the logarithm base 2 of both sides:
(2 A)/3 = (log(20/13))/(log(2))

Multiply both sides by 3/2:

A = (3 ln(20/13))/(2 ln(2)= 0.932 x 10^9 =9.32 x 10^8 - Years - the original age of the rock.

b. Analysis of another rock shows that it contains15​% of its original​ element; the other 85​% decayed into lead. How old is the​ rock?

Let the original age of the rock =A

15% =100% x 2^-(A/1.5), solve for A

0.15 = 2^(-0.666667 A)

0.15 = 3/20 and 2^(-0.666667 A) = 2^(-2 A/3):
3/20 = 2^(-(2 A)/3)

3/20 = 2^(-(2 A)/3) is equivalent to 2^(-(2 A)/3) = 3/20:
2^(-(2 A)/3) = 3/20

Take reciprocals of both sides:
2^((2 A)/3) = 20/3

Take the logarithm base 2 of both sides:
(2 A)/3 = (log(20/3))/(log(2))

Multiply both sides by 3/2:
A = (3 ln(20/3))/(2 ln(2))=4.1 x 10^9 Years - original age of the rock

Dec 6, 2017