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.
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
