### 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 d times 2 = 37.44 d

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 times 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 times 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 times 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: U-238 has a half-life of 4.46 x 109 years. Estimates of the age of the universe range from 9 x 109 years to 23 x 109 years (Cauldrons in the Cosmos: Nuclear Astrophysics, C.E. Rolfs and W.S. Rodney, Univ. of Chicago, 1988, p. 477). What fraction of this isotope present at the start of the universe remains today? Calulate for both minimum and maximum values, as well as a median value of 16 x 109 years.

Solution:

1) Calculation for the median value:

(16 x 109) / (4.46 x 109) = 3.587 half-lives

2) What fraction remains?

(1/2)3.587 = 0.0832

8.32% remains

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