# Equation half life dating could him

### 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 10^{23} 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 10

^{23}) (0.0213) = 1.28 x 10^{22}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:

2.00 mg / 128.0 mg = 0.015625How many half-lives must have elaspsed to get to 0.015625 remaining?

(1/2)24 days / 6 half-lives = 4.00 days (the length of the half-life)^{n}= 0.015625

n log 0.5 = log 0.015625

n = log 0.5 / log 0.015625

n = 6Video: An Alternate Solution to the Above Problem

Problem #5: U-238 has a half-life of 4.46 x 10^{9} years. How much U-238 should be present in a sample 2.5 x 10^{9} years old, if 2.00 grams was present initially?

Solution:

(2.5 x 10^{9}) / (4.46 x 10^{9}) = 0.560 (the number of half-lves that have elapsed)(1/2)

^{0.560}= 0.678 (the decimal fraction of U-238 remaining)2.00 g x 0.678 = 1.36 g remain

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}= 0.01n 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-livesOne day = one half-life; (1/2)

^{1.0042}= 0.4985465 remaining = 4.98 gTwo days = two half-lives; (1/2)

^{2.0084}= 0.2485486 remaining = 2.48 gSeven 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}= 0.05n 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}= 0.25n = 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}= 0.25n log 0.5 = log 0.25

n = log 0.25 / log 0.5

n = 2

Go to introductory half-life discussion

Probs 11-25

Probs 26-40

Problems involving carbon-14

Return to Radioactivity menu