Half-Life Problems #11 - 25

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 decayed

1 − 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% remaining

1% = 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 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: 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 #24: What precentage of the parent isotope remains after 0.5 half lives have passed?

Solution:

(1/2)ⁿ = decimal amount remaining

where 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 = 4

4 half-lives = 0.0625 remaining

0.0625 g