Half-Life Problems #26 - 40 (incomplete)

Go to introductory half-life discussion

Got to half-life problems #1 - 10

Go to half-life problems #11-25

Go to half-life problems involving carbon-14

Return to Radioactivity menu


Problem #26: The half life in two different samples, A and B, of radio-active nuclei are related according to T(1/2,B) = T(1/2,A)/2. In a certain period the number of radio-active nuclei in sample A decreases to one-fourth the number present initially. In the same period the number of radio-active nuclei in sample B decreases to a fraction f of the number present initially. Find f.

Solution:

a) Sample A underwent two half-lives:

1 ---> 1/2 ---> 1/4

b) Let us set the length of one half-life of A equal to 1. Therefore the total amount of elapsed time for A was 2.

T(1/2,B) = T(1/2,A)/2

Since T(1/2,A) = 1,

we now know that T(1/2,B) = 1/2

c) Allow B to go through several half-lives such that the total amount of time = 2. This is 4 half-lives (1/2 + 1/2 + 1/2 + 1/2 = 2):

Four half-lives:

1 ---> 1/2 ---> 1/4 ---> 1/8 ---> 1/16

f = 1/16


Problem #27: You have 20.0 grams of 32-P that decays 5% daily. How long will it take for half the original to decay?

Solution:

In 24 hours, the sample goes from 100% to 95%

(1/2)n = 0.95

n log 0.5 = log 0.95

n = 0.074

24 hrs / 0.074 = 324 hrs (one half-life)


Problem #28: A sample of radioactive isotopes contains two different nuclides, labeled A and B. Initially, the sample composition is 1:1, i.e., the same number of nuclei A as nuclei B. The half-life of A is 3 hours and, that of B, 6 hours. What is the expected ratio A/B after 18 hours?

Solution:

A has a half-life of 3 hrs, so 18 hrs = 6 half-lives.
B has a half-life of 6 hrs, so 18 hrs = 3 half-lives.

After 6 half lives, the fraction of A left is 1/(26) = 1/64
The fraction of B left is 1/(23) = 1/8.

Since A/B started out at 1/1, A/B at 18 hrs = (1/64) / (1/8) = 1/8.

This could also be expressed as: 0.56 / 0.53 = 0.56-3 = 1/8


Problem #29: If the half-life of 238-U is 4.5 x 109 y and the half-life of 235-U is 7.1 x 108 y and the age of the Earth is 4.5 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:

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

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

4.5 x 109 y / 4.5 x 109 y = one half-life elapsed

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

Initially present was 99.3 + 99.3 = 198.6 g

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

4.5 x 109 y / 7.1 x 108 y = 6.338 half-lives elapsed

(1/2)6.338 = 0.01236 (the decimal fraction remaining after 6.338 half-lives)

0.7 g / 0.01236 = 56.5 g initially present

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

238-U ⇒ 198.6 / 255.2 = 77.82%
235-U ⇒ 56.5 / 255.2 = 22.18% (I did this one by subtraction from 100%.)

Problem #30: The isotope Ra-226 decays to Pb-206 in a number of stages which have a combined half-life of 1640 years. Chemical analysis of a certain chunk of concrete from an atomic-bombed city, preformed by an archaeologist in the year 6264 AD, indicated that it contained 2.50 g of Ra-226 and 6.80 g of Pb-206. What was the year of the nuclear war?

Solution:

Start by ignoring a few chemical realities and assume all the Ra-226 ends up as lead.

a) Calculate moles of Ra-226 decayed:

6.80 g / 205.974465 g/mol = 0.033013801 mol of Pb-206 decayed

∴ 0.033013801 mol of Ra-226 decayed

b) Calculate grams of Ra-226 initially present:

(0.033013801 mol) (226.02541 g/mol) = 7.462 g of Ra-226 decayed

7.462 + 2.50 = 9.962 g of Ra-226 initially present

c) Calculate decimal fraction of Ra-226 remaining:

2.50 g / 9.962 g = 0.251

d) Calculate number of half-lives elapsed:

(1/2)n = 0.251

n = 2

e) Calculate year of war:

1640 x 2 = 3280 y elapsed since war

6264 - 3280 = 2984 AD


Problem #31: A radioactive sample contains 3.25 x 1018 atoms of a nuclide that decays at a rate of 3.4 x 1013 disintegrations per 26 min.

(a) What percentage of the nuclide will have decayed after 159 days?
(b) What is the half-life of the nuclide?

Solution to a:

159 days x 24 hrs/day x 60 min/hour = 228960 min

228960 min x (3.4 x 1013 disintegrations per 26 min) = 2.994 x 10 x 1017 total dis in 159 days

2.994 x 10 x 1017 / 3.25 x 1018 = 0.0921

9.21% has disintegrated

Solution to b:

0.9079 is the decimal fraction of the substance remaining since 0.0921 has gone away

(1/2)n = 0.9079

n log 0.5 = log 0.9079

n = 0.139 half-lives

159 day / 0.139 = 1144 days


Problem #32: The radioisotope potassium-40 decays to argon-40 by positron emission with a half life of 1.27 x 109 yr. A sample of moon rock was found to contain 78 argon-40 atoms for every 22 potassium-40 atoms. The age of the rock is . . .

Solution:

Assume the sample was 100% K-40 at start. In the present day, the sample contains 78% Ar-40 and 22% K-40. We will use 0.22, the decimal percent of K-40 remaining:

(1/2)n = 0.22

where n is the number of half-lives.

n log 0.5 = log 0.22

n = 2.18

What is the total elapsed time?

(2.18) (1.27 x 109) = 2.77 x 109 yrs

Problem #33: What is the age of a rock in which the mass ratio of Ar-40 to K-40 is 3.8? K-40 decays to Ar-40 with a half-life of 1.27 x 109 yr.

Solution:

Since, the sample is 3.8 parts by mass Ar and 1 part K, the orginal sample contained 4.8 parts K and zero parts Ar.

What is the decimal amount of K-40 that remains?

1 part divided by 4.8 parts = 0.20833

How many half-lives are required to reach 0.20833 remaining?

(1/2)n = 0.20833

where n is the number of half-lives.

n log 0.5 = log 0.20833

n = 2.263

What is the total elapsed time?

(2.263) (1.27 x 109) = 2.87 x 109 yrs

Problem #34: The half-life for the following process is 4.5 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. You might want to try using those numbers as you study the following explanation.

Solution:

Determine millimoles of Pb-206:

14.5 mg / 206 mg/mmol = 0.07039 mmol

Determine mg of U-238 that must have decayed:

0.07309 mmol times 238 mg/mmol = 16.75 mg of U-238

Total U-238 present at start of decay:

43.20 mg + 16.75 mg = 59.95 mg

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

Determine how much time has elapsed:

4.5 x 109 yr times 0.47273 = 2.1 x 109 yr (to two sig figs)

Problem #35: 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 4.5 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 times 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 times 4.5 x 109 = 2.64 x 109 yr

Problem #36: Natural samarium (average atomic mass 150.36) contains 14.99% of the radioactive isotope Sm-147. A 1 g sample of natural Sm has an activity of 89 decays per second. Estimate the half life of Sm-147.

Solution:

1 g times 0.1499 = 0.1499 g of Sm-147

0.1499 g / 146.915 g/mol = 0.00102032 mol

0.00102032 mol times 6.022 x 1023 mol¯1 = 6.144367 x 1020 atoms

6.144367 x 1022 atoms / 2 = 3.0721835 x 1020 atoms

3.0721835 x 1020 atoms / 89 decays/sec = 3.45189 x 1018 sec

3.45189 x 1018 sec = 1.094 x 1011 yr.

The Wiki article for isotopes of samarium gives 1.06 x 1011 yr.

Implicit in this solution is that the decays rate of 89 decays/second remains constant for the entire half-life. The decay rate actually becomes lesser over time but, for purposes of making an estimate, we ignore this.


Problem #37: 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. I found that value here.

1.3 x 108 yr divided by 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 times 205.97 g/mol = 0.0000573653 g

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


Go to introductory half-life discussion

Got to half-life problems #1 - 10

Go to half-life problems #11-25

Go to half-life problems involving carbon-14

Return to Radioactivity menu