Half-Life Problems: Part Two

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Problem #16: Rn-222 has a half-life of 3.82 days. How long before only 1/16 of the original sample remains?

Solution:

recognize 1/16 as a fraction associated with 4 half-lives (from (1/2)4 = 1/16)

3.82 days x 4 = 15.3 days


Problem #17: U-238 has a half-life of 4.46 x 109 years. Estimates of the age of the universe range from 9 x 109 years to 23 x 109 years (Cauldrons in the Cosmos: Nuclear Astrophysics, C.E. Rolfs and W.S. Rodney, Univ. of Chicago, 1988, p. 477). 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) / (4.46 x 109) = 3.587 half-lives

2) What fraction remains?

(1/2)3.587 = 0.0832

8.32% remains


Problem #18: A sample of Se-83 registers 1012 disintegrations per second when first tested. What rate would you predict for this sample 3.5 hours later, if the half-life is 22.3 minutes?

Solution:

210 min / 22.3 min = 9.42 half-lives (210 min is 3.5 hours)

(1/2)9.42 = 0.00146 (the decimal fraction remaining)

1012 x 0.00146 = 1.46 x 109 disintegrations per second remaining


Problem #19: Iodine-131 has a half-life of 8.040 days. If we start with a 40.0 gram sample, how much will remain after 24.0 days?

Solution:

24.0 days / 8.040 days = 2.985 half-lives

(1/2)2.985 = 0.1263 (the decimal fraction remaining)

40.0 g x 0.1263 = 5.05 g


Problem #20: If you start with 2.97 x 1022 atoms of molybdenum-99 (half-life = 65.94 hours), how many atoms will remain after one week?

Solution:

one week = 168 hours

168 / 65.94 = 2.548

(1/2)2.548 = 0.171 (the decimal fraction remaining)

(2.97 x 1022) x 0.171 = 5.08 x 1021


Problem #21: The isotope H-3 has a half life of 12.26 years. Find the fraction remaining after 49 years.

Solution:

49 / 12.26 = 3.9967

(1/2)3.9967 = 0.0626


Problem #22: How long will it take for a 64.0 g sample of Rn-222 (half-life = 3.8235 days) to decay to 8.00 g?

Solution:

8.00 / 64.0 = 0.125 (the decimal fraction remaining)

(1/2)n = 0.125

by experience, n = 3 (remember that 0.125 is 1/8)

3.8235 x 3 = 11.4705 days


Problem #23: 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 #24: 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 #25: 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 #26: 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 #27: 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 #28: Manganese-56 is a beta emitter with a half-life of 2.6 h. What is the mass of manganese-56 in a 1.0 g sample of the isotope at the end of 10.4 h?


Problem #29: 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 #30: 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 #31: 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 #32: A scientist needs 10.0 micrograms of Ca-47 (half-life = 4.50 days) to do an experiment on an animal. If the delivery time is 50.0 hours, how many micrograms of 47CaCO3 must the scientist order?

Solution:

4.50 days x 24 hrs/day = 108 hrs

50/108 = 0.463 half-lives

(1/2)0.463 = 0.725 (the decimal portion of Ca-47 remaining after 50 hrs)

10.0 mg / 0.725 = 13.8 mg


Problem #33: What precentage of the parent isotope remains after 0.5 half lives have passed?

Solution:

(1/2)n = decimal percent remaining

where n = the number of half-lives

(1/2)0.5 = 0.707

70.7%


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

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