Half-Life Problems #1 - 10

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Problem #1: The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes has elapsed?

Solution:

7.2 / 2.4 = 3 half-lives

(1/2)3 = 0.125 (the amount remaining after 3 half-lives)

100.0 g x 0.125 = 12.5 g remaining


Problem #2: Pd-100 has a half-life of 3.6 days. If one had 6.02 x 1023 atoms at the start, how many atoms would be present after 20.0 days?

Solution:

20.0 / 3.6 = 5.56 half-lives

(1/2)5.56 = 0.0213 (the decimal fraction remaining after 5.56 half-lives)

(6.02 x 1023) (0.0213) = 1.28 x 1022 atoms remain


Problem #3: Os-182 has a half-life of 21.5 hours. How many grams of a 10.0 gram sample would have decayed after exactly three half-lives?

Solution:

(1/2)3 = 0.125 (the amount remaining after 3 half-lives)

10.0 g x 0.125 = 1.25 g remain

10.0 g − 1.25 g = 8.75 g have decayed

Note that the length of the half-life played no role in this calculation. In addition, note that the question asked for the amount that decayed, not the amount that remaning.


Problem #4: After 24.0 days, 2.00 milligrams of an original 128.0 milligram sample remain. What is the half-life of the sample?

Solution:

The decimal fraction remaining:

2.00 mg / 128.0 mg = 0.015625

2) How many half-lives must have elaspsed to get to 0.015625 remaining?

(1/2)n = 0.015625

n log 0.5 = log 0.015625

n = log 0.5 / log 0.015625

n = 6

3) Determine the half-life:

24 days / 6 half-lives = 4.00 days

Video: An Alternate Solution to the Above Problem


Problem #5: A radioactive isotope decayed to 17/32 of its original mass after 60 minutes. Find the half-life of this radioisotope.

Solution:

17/32 = 0.53125 (this is the decimal amount that remains)

(1/2)n = 0.53125

n log 0.5 = log 0.53125

n = 0.91254 (this is how many half-lives have elapsed)

60 min / 0.91254 = 65.75 min

n = 66 min (to two sig figs)


Problem #6: How long will it take for a 40.0 gram sample of I-131 (half-life = 8.040 days) to decay to 1/100 its original mass?

Solution:

(1/2)n = 0.01

n log 0.5 = log 0.01

n = 6.64

6.64 x 8.040 days = 53.4 days


Problem #7: Fermium-253 has a half-life of 0.334 seconds. A radioactive sample is considered to be completely decayed after 10 half-lives. How much time will elapse for this sample to be considered gone?

Solution:

0.334 x 10 = 3.34 seconds

Problem #8: At time zero, there are 10.0 grams of W-187. If the half-life is 23.9 hours, how much will be present at the end of one day? Two days? Seven days?

Solution:

24.0 hr / 23.9 hr/half-life = 1.0042 half-lives

One day = one half-life; (1/2)1.0042 = 0.4985465 remaining = 4.98 g

Two days = two half-lives; (1/2)2.0084 = 0.2485486 remaining = 2.48 g

Seven days = 7 half-lives; (1/2)7.0294 = 0.0076549 remaining = 0.0765 g


Problem #9: 100.0 grams of an isotope with a half-life of 36.0 hours is present at time zero. How much time will have elapsed when 5.00 grams remains?

Solution:

5.00 / 100.0 = 0.05 (decimal fraction remaining)

(1/2)n = 0.05

n log 0.5 = log 0.05

n = 4.32 half-lives

36.0 hours x 4.32 = 155.6 hours


Problem #10: How much time will be required for a sample of H-3 to lose 75% of its radioactivity? The half-life of tritium is 12.26 years.

Solution:

If you lose 75%, then 25% remains. Use 0.25 rather than 25%.

(1/2)n = 0.25

n = 2 (remember (1/2)2 = 1/4 and 1/4 = 0.25)

12.26 x 2 = 24.52 years

Comment: the more general explanation follows:

(1/2)n = 0.25

n log 0.5 = log 0.25

n = log 0.25 / log 0.5

n = 2


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