### Half-life problems involving uranium-238

Problem #56: U-238 has a half-life of 4.468 x 109 years. How much U-238 should be present in a sample 2.50 x 109 years old, if 2.00 grams was present initially? (I found the U-238 half-life value here.)

Solution:

(2.5 x 109) / (4.468 x 109) = 0.55953 (the number of half-lives that have elapsed)

(1/2)0.55953 = 0.678523 (the decimal fraction of U-238 remaining)

(2.00 g) (0.678523) = 1.36 g (to three sig figs)

Problem #57: A sample of rock is known to contain the isotopes U-238 and Pb-206 in the mass ratio of 2:1. What is the age of the sample assuming all the Pb-206 has originated from the decay of U-238? The half-life of U-238 is 4.468 x 109 years.

Solution:

1) Determine moles of U-238 and Pb-206:

Assume 200. g of U-238 and 100. g of Pb-206.

moles U-238 ---> 200. g / 238.0508 g/mol = 0.84016 mol
moles Pb-206 ---> 100. g / 205.9745 g/mol = 0.48550 mol

At time zero, there was 1.32566 mole of U-238 and zero mole of Pb-206.

2) The percentage of U that remains in the present is this:

0.84016 / 1.32566 = 0.633767

3) Determine how many half-lives have elapsed:

(1/2)n = 0.633767

n log 0.5 = log 0.633767

n = 0.65798

4) Determine how much time has elapsed:

(0.65798) (4.468 x 109 yr) = 2.94 x 109 yr

5) Here's another solution to the above problem.

Problem #58: A sample of rock contains 64.27 mg of U-238 (isotopic mass 238.0508 amu) and 22.66 mg of Pb-206 (isotopic mass 205.9745 amu). Assuming that radioactive decay of U-238 is the only source of Pb-206, estimate the rock's age. Half life of U-238 radioactive disintegration is = 4.468 x 109 years.

Solution:

1) Determine moles of U-238 and Pb-206:

We could assume grams rather than milligrams, but let's not.

moles U-238 ---> 0.06427 g / 238.0508 g/mol = 0.00026998 mol
moles Pb-206 ---> 0.02266 g / 205.9745 g/mol = 0.00011001 mol

At time zero, there was 0.00037999 mole of U-238 and zero mole of Pb-206.

2) The percentage of U that remains in the present is this:

0.00026998 / 0.00037999 = 0.7104924

3) Determine how many half-lives have elapsed:

(1/2)n = 0.7104924

n log 0.5 = log 0.7104924

n = 0.493732

4) Determine the age of the rock:

(0.493732) (4.468 x 109 yr) = 2.206 x 109 yr

Problem #59: A rock is found contains uranium-238 and also lead-206. Scientist analyze the rock for these two elements and find that the total mass of uranium in the rock is 2.40 g, while the amount of lead is 1.11 g. How old is this rock?

Solution:

1) Determine moles of U-238 and Pb-206:

Assume 240. g of U-238 and 111 g of Pb-206.

moles U-238 ---> 240. g / 238.0508 g/mol = 1.00819 mol
moles Pb-206 ---> 111 g / 205.9745 g/mol = 0.53890 mol

At time zero, there was 1.54709 mole of U-238 and zero mole of Pb-206.

2) The percentage of U that remains in the present is this:

1.00819 / 1.54709 = 0.6516686

3) Determine how many half-lives have elapsed:

(1/2)n = 0.6516686

n log 0.5 = log 0.6516686

n = 0.61779

4) Determine how much time has elapsed:

(0.61779) (4.468 x 109 yr) = 2.76 x 109 yr

5) Here's another solution to the above problem.

Problem #60: If the half-life of 238-U is 4.468 x 109 y and the half-life of 235-U is 7.04 x 108 y and the age of the Earth is 4.468 x 109 y and if the percentage of 238-U in the Earth is 99.3% and 235-U is 0.7% then what were their percentages when the Earth was formed?

Solution:

1) Assume 100 g of present-day uranium is present. In it, there are 99.3 g of 238-U and 0.7 g of 235-U

2) Determine amount of 238-U when Earth formed:

4.468 x 109 y / 4.468 x 109 y = one half-life elapsed

99.3 g represents the amount of uranium present after one half-life.

Initially present was 99.3 + 99.3 = 198.6 g

3) Determine amount of 235-U when Earth formed:

4.468 x 109 y / 7.04 x 108 y = 6.3466 half-lives elapsed

(1/2)6.3466 = 0.012288 (the decimal fraction remaining after 6.3466 half-lives)

0.7 g / 0.012288 = 56.97 g initially present

4) Calculate percentages for each isotope when the Earth was formed:

198.6 + 56.97 = 255.57

238-U ⇒ 198.6 / 255.57 = 77.7%
235-U ⇒ 56.97 / 255.57 = 22.3% (I did this one by subtraction from 100%.)

Problem #61: The half-life for the following process is 4.468 x 109 yr.

U-238 ---> Pb-206

A mineral sample contains 43.20 mg of U-238 and 14.50 mg of Pb-206. What is the age of the mineral?

Here is another set of numbers for this problem: 40.60 mg of U-238 and 12.80 mg of Pb-206. Use those numbers to do a parallel calculation as you study the following explanation.

Solution:

1) Determine millimoles of Pb-206:

14.5 mg / 206 mg/mmol = 0.07039 mmol

2) Determine mg of U-238 that must have decayed:

(0.07309 mmol) (238 mg/mmol) = 16.75 mg of U-238

3) Total U-238 present at start of decay:

43.20 mg + 16.75 mg = 59.95 mg

4) Determine how many half-lives have elapsed:

43.20 / 59.95 = 0.7206 (this is the decimal fraction of U-238 remaining)

(1/2)n = 0.7206 (where n = the number of half-lives)

n log 0.5 = log 0.7206

n = 0.47273

5) Determine how much time has elapsed:

(4.468 x 109 yr) (0.47273) = 2.112 x 109 yr (to four sig figs)

Problem #62: A rock from Australia was found to contain 0.435 g of Pb-206 to every 1.00 g of U-238. Assuming that the rock did not contain any Pb-206 at the time of its formation, how old is the rock? The half life of U-238 is 4.468 x 109 years.

Solution:

1) Let's get the initial amount of U-238:

0.435 g / 205.974 g/mol = 0.002111 mol of Pb-206

There is a 1:1 molar ratio between U-238 decaying and Pb-206 forming.

(0.002111 mol) (238.051 g/mol) = 0.502525661 g

Let's use 0.5025 g

initial amount of U = 1.5025 g

2) Decimal amount of U remaining in the present:

1.00 / 1.5025 = 0.6656

3) Find number of half-lives elapsed:

(1/2)n = 0.6656

n log 0.5 = log 0.6656

n = 0.587 half-lives

4) Time elapsed:

(0.587) (4.468 x 109) = 2.62 x 109 yr

Problem #63: How much Pb-206 will be in a rock sample that is 1.3 x 108 years old and that contains 3.25 mg of U-238?

Solution:

The half-life of U-238 is 4.468 x 109 yr.

1.3 x 108 yr / 4.468 x 109 yr = 0.029098 (this is how many half-lives have elapsed)

(1/2)0.029098 = 0.9800 (this is the decimal amount of U-238 remaining)

3.25 mg is to 0.98 as x is to 1

x = 3.3163 mg (this is the amount of U-238 at the start of the decay)

3.3163 − 3.25 = 0.0663 mg (of U-238 that decayed)

0.0000663 g / 238.05 g/mol = 2.78513 x 10-7 mol of U-238

This means 2.78513 x 10-7 mol of Pb-206 was formed.

(2.78513 x 10-7 mol) (205.97 g/mol) = 0.0000573653 g

to two sig figs and using mg, this is 0.057 mg

Problem #64: U-238 has a half-life of 4.468 x 109 years. Estimates of the age of the universe range from 9 x 109 years to 23 x 109 years. What fraction of this isotope present at the start of the universe remains today? Calulate for both minimum and maximum values, as well as a median value of 16 x 109 years.

Solution:

1) Calculation for the median value:

(16 x 109) / (4.468 x 109) = 3.581 half-lives

2) What fraction remains?

(1/2)3.581 = 0.08356

8.356% remains

Problem #65: Estimate the age of rocks in which the mole ratio of U-238 to Pb-205 is 0.75. The half life of U-238 is 4.468 x 109 yr.

Solution:

1) The U to Pb mole ratio is 0.75 to 1 in the present:

Therefore, in the past, at time zero, there was 1.75 mole of U to zero mole of Pb.

2) The percentage of U that remains in the present is this:

0.75 / 1.75 = 0.42857 (extra digits for the moment)

3) Determine how many half-lives have elapsed:

(1/2)n = 0.42857

n log 0.5 = log 0.42857

n = 1.2224

4) Determine how much time has elapsed:

(1.2224) (4.468 x 109 yr) = 5.462 x 109 yr