Half-Life Problems #26 - 40

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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 P-32 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: The ratio of tritium, H-3, to hydrogen, H-1, in a sample of water was 1:1x1019. If the half life of tritium is 12.25 years, calculate the actual number of tritium atoms remaining in 10.0 g water after a period of 49 years.

Solution:

1) Moles, then molecules of water:

10.0 g / 18.0148 g/mol = 0.555099 mol

(0.555099 mol) (6.02214 x 1023 molecules / mol) = 3.342884 x 1023 molecules

2) Atoms of hydrogen in 10.0 g of water:

(2 atoms/molecules) (3.342884 x 1023 molecules) = 6.685768 x 1023 atoms

3) Atoms of H-3:

6.685768 x 1023 / 1 x 1019 = 66858 atoms of H-3 (to the nearest whole number)

4) Half-lives elapsed:

49 yr / 12.25 yr = 4

5) Amount remaining after 4 half-lives:

(1/2)4 = 1/16 = 0.0625

6) atoms remaining after 4 half-lives:

(66858 atoms) (0.0625) = 4179 atoms (to the nearest whole number)

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. What is the age of the rock?

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: A radioactive isotope has a half-life of 4.5 days. What fractions of the sample will exist after 9 and 18 days?

(a) 1/2 and 1/4 of the original amount
(b) 1/9 and 1/18 of the original amount
(c) 1/4 and 1/16 of the original amount
(d) 1/4 and 1/8 of the original amount

Solution:

Half-lives elapsedzeroonetwothreefour
Days elapsed04.5913.58
Fraction remaining11/21/41/81/16

  Answer: (c) 1/4 and 1/16 of the original


Problem #35: Selenium-75 has a half-life of 120 days and is used medically for pancreas scans. (a) Approximately how much selenium-75 would remain of a 0.050 g sample that has been stored for one year? (b) How long would it take for a sample of selenium-75 to lose 99% of its radioactivity?

Solution to (a):

365 day / 120 day = 3.0417 half-lives

(1/2)3.0417 = 0.12144 (this is the decimal amount that remains)

0.12144 x 0.050 g = 0.006072 g

to two sig figs = 0.0061 g

Solution to (b):

99% gone means 1% remaining, which is 0.01 as a decimal

(1/2)n = 0.01

n log 0.5 = log 0.01

n = 6.643856 (this is the number of half-lives elapsed)

120 day times 6.643856 = 797 days


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: You measure the radioactivity of a substance, then when measuring it 120 days later, you find that it only has 54.821% of the radioactivity it had when you first measured it. What is the half life of that substance?

Solution:

1) How many half-lives have elapsed in the 120 days?

(1/2)n = 0.54821

n log 0.5 = log 0.54821

n = 0.8672

2) Determine the half-life.

120 day / 0.8672 = 138.4 day

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-lives

2.30 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 15 grams 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 remaining

n 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 1 as x is to 0.251188

x = 3.8 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-second (3.125%) of its original value?

Solution:

(1/2)n = 0.03125

n log 0.5 = log 0.03125

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

5.27 x 5 = 26.35 years

You may convert years to hours.


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