Go to introductory half-life discussion

**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)/2Since 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}= 0.95n 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/(2

^{6}) = 1/64

The fraction of B left is 1/(2^{3}) = 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: 0.5

^{6}/ 0.5^{3}= 0.5^{6-3}= 1/8

**Problem #29:** If the half-life of 238-U is 4.5 x 10^{9} y and the half-life of 235-U is 7.1 x 10^{8} y and the age of the Earth is 4.5 x 10^{9} 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 10^{9}y / 4.5 x 10^{9}y = one half-life elapsed99.3 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 10^{9}y / 7.1 x 10^{8}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 decayed7.462 + 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}= 0.251n = 2

e) Calculate year of war:

1640 x 2 = 3280 y elapsed since war6264 - 3280 = 2984 AD

**Problem #31:** A radioactive sample contains 3.25 x 10^{18} atoms of a nuclide that decays at a rate of 3.4 x 10^{13} 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 min228960 min x (3.4 x 10

^{13}disintegrations per 26 min) = 2.994 x 10 x 10^{17}total dis in 159 days2.994 x 10 x 10

^{17}/ 3.25 x 10^{18}= 0.09219.21% has disintegrated

**Solution to b:**

0.9079 is the decimal fraction of the substance remaining since 0.0921 has gone away(1/2)

^{n}= 0.9079n 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 10^{9} 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 0.22n = 2.18

What is the total elapsed time?

(2.18) (1.27 x 10^{9}) = 2.77 x 10^{9}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 10^{9} 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 0.20833n = 2.263

What is the total elapsed time?

(2.263) (1.27 x 10^{9}) = 2.87 x 10^{9}yrs

**Problem #34:** The half-life for the following process is 4.5 x 10^{9} 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 10^{9}yr times 0.47273 = 2.1 x 10^{9}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 10^{9} years.

**Solution:**

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

0.435 g / 205.974 g/mol = 0.002111 mol of Pb-206There 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}= 0.6656n log 0.5 = log 0.6656

n = 0.587 half-lives

4) time elapsed:

0.587 times 4.5 x 10^{9}= 2.64 x 10^{9}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-1470.1499 g / 146.915 g/mol = 0.00102032 mol

0.00102032 mol times 6.022 x 10

^{23}mol¯^{1}= 6.144367 x 10^{20}atoms6.144367 x 10

^{22}atoms / 2 = 3.0721835 x 10^{20}atoms3.0721835 x 10

^{20}atoms / 89 decays/sec = 3.45189 x 10^{18}sec3.45189 x 10

^{18}sec = 1.094 x 10^{11}yr.The Wiki article for isotopes of samarium gives 1.06 x 10

^{11}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 10^{8} years old and that contains 3.25 mg of U-238?

**Solution:**

The half-life of U-238 is 4.468 x 10^{9}yr. I found that value here.1.3 x 10

^{8}yr divided by 4.468 x 10^{9}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-238This 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 gto two sig figs and using mg, this is 0.057 mg

**Problem #38:** Arsenic-74 is a medical radioisotopes with a half-life of 18 days. If the initial amount of arsenic-74 injected is 2.30 mCi, how much arsenic-74 is left in the body after 54 days?

Solution:

54 days / 18 days = 3 half-lives elapsed(1/2)

^{3}= 0.125 <--- amount remaining after three half-lives2.30 mCi x 0.125 = 0.2875 mCi

to three sig figs, 0.288 mCi

**Problem #39:** Assume that today there is 10 grams of substance, while 1000 years ago there was 100 grams of it. If there is 15grams of substance today, how much will there be 600 years from now?

**Solution:**

1) What is needed is the length of the half-life.

10 g / 100 g = 0.1 <--- this is the decimal amount remaining after 1000 years(1/2)

^{n}= 0.1 <--- where n is the number of half-lives needed to elapse to give 0.1 remainingn log 0.5 = log 0.1

n = 3.321928 half-lives

1000 yr / 3.321928 = 301.03 yr

2) Now, we turn to the 15 g sample. We need to know how many half-lives have elapsed

600 yr / 301.03 yr = 1.99316 half-lives(1/2)

^{1.99316}= 0.251188 <--- that's the decimal amount remaining after 600 years.15 g is to 0.251188 as x is to 1

x = 59.716 g

**Problem #40:** A radionuclide, cobalt-60, has a half-life of 5.27 years. How many hours would it take for the activity to diminish to one-thirty-secondth (3.125%) of its original value?

**Solution:**

(1/2)^{n}= 0.03125n log 0.5 = log 0.03125

n = 5 (this is how many half-lives have elapsed)

5.27 x 5 = 26.35 years

I'll let you convert years to hours.