Ten Examples | Problems involving carbon-14 | |
Probs 1-10 | Problems involving uranium-238 | |
Probs 26-40 | Examples and Problems only (no solutions) | Return to Radioactivity menu |
Problem #11: The half life of iodine-131 is 8.040 days. What percentage of an iodine-131 sample will remain after 40.20 days?
Solution:
40.20 d / 8.040 d = 5(1/2)5 = 0.03125
percent remaining = 3.125%
Problem #12: The half-life of thorium-227 is 18.72 days How many days are required for three-fourths of a given amount to decay?
Solution:
3/4 = 0.75 <--- amount decayed1 − 0.75 = 0.25 <--- amount remaining
(1/2)n = 0.25
n = 2
(18.72 day) (2) = 37.44 day
Problem #13: If you start with 5.32 x 109 atoms of Cs-137, how much time will pass before the amount remaining is 5.20 x 106 atoms? The half-life of Cs-137 is 30.17 years.
Solution:
5.20 x 106 / 5.32 x 109 = 0.0009774436 (the decimal amount remaining)(1/2)n = 0.0009774436
n log 0.5 = log 0.0009774436
n = 9.99869892 half-lives
(30.17 yr) (10) = 301.7 yr
Problem #14: The half-life of the radioactive isotope phosphorus-32 is 14.3 days. How long until a sample loses 99% of its radioactivity?
Solution:
99% loss means 1% remaining1% = 0.01
(1/2)n = 0.01
n log 0.5 = log 0.01
n = 6.643856
(14.3 day) (6.643856) = 95.0 day
Problem #15: The half-life of palladium-100 is 4 days. After 12 days a sample of Pd-100 has been reduced to a mass of 4.00 mg. (a) Determine the starting mass. (b) What is the mass 8 weeks after the start?
Solution:
12 day / 4 day = 3(1/2)3 = 0.125
4.00 mg / 0.125 = 32.0 mg
8 weeks = 56 days
56 d / 4 = 14 half-lives
(1/2)14 = 0.000061035
(32.0 mg) (0.000061035) = 0.00195 mg (rounded to three figs)
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 1/24 = 1/16)3.82 days x 4 = 15.3 days
Problem #17: One-eighth of a radioactive sample remains 9 days after it was brought into the lab. What is the half-life?
Solution:
One-eighth is evocative of three half-lives.9 day / 3 = 3 day
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 hours168 / 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: 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 hrs50/108 = 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 #24: What precentage of the parent isotope remains after 0.5 half lives have passed?
Solution:
(1/2)n = decimal amount remainingwhere n = the number of half-lives
(1/2)0.5 = 0.707
The question asks for a percentage, so 70.7%
Problem #25: Manganese-56 has 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?
Solution:
10.4 / 2.6 = 44 half-lives = 0.0625 remaining
0.0625 g
Ten Examples | Problems involving carbon-14 | |
Probs 1-10 | Problems involving uranium-238 | |
Probs 26-40 | Examples and Problems only (no solutions) | Return to Radioactivity menu |