Download
1 / 4
40 likes | 147 Views
HIGHER GRADE CHEMISTRY CALCULATIONS. Half-life is the time taken for half of a radioisotope to decay. This value is constant for that particular radioisotope. Usually questions about radioactivity involve Half-life The time over which the radioactivity has been measured
E N D
HIGHER GRADE CHEMISTRY CALCULATIONS Half-life is the time taken for half of a radioisotope to decay. This value is constant for that particular radioisotope. Usually questions about radioactivity involve • Half-life • The time over which the radioactivity has been measured • The quantity or intensity of the radiation. Radioactivity and Half-Life Worked example 1. A radioisotope has a half-life of 4 days. The radioisotope had an initial count rate of 800 counts per minute. What will be the count rate after 16 days? At start count rate = 800 counts min-1. After 4 days = 400 counts min-1. After 8 days = 200 counts min-1. After 12 days = 100 counts min-1. After 16 days = 50 counts min-1.
Worked example 2. • Thallium-208 has a half life of 3.1 minutes and decays by beta emission to form a stable isotope. • What mass of a 2.08 g sample of Tl-208 will remain unchanged after • 9.3 minutes? • (b) How many atoms will have decayed. • (c) Identify the stable isotope formed by the decay of Tl-208 by beta emission. (a) At start mass = 2.08g After 3.1 days = 1.04g After 6.2 days = 0.52g After 9.3 days = 0.26g (b) At start no. of atoms = 2.08/208 x 6.02 x 1023 = 6.02 x 1021 After 3.1 days no. of atoms of Tl-208 left undecayed= 3.01 x 1021 After 6.2 days no. of atoms of Tl-208 left undecayed= 1.505 x 1021 After 9.3 days no. of atoms of Tl-208 left undecayed= 0.7525 x 1021 No. of atoms of Tl-208 which have decayed= 6.02 x 1021 -0.7525 x 1021. 5.2675 x 1021 Higher Grade Chemistry (c) Pb-208 has been formed.
(b) No of atoms at start = 3.01 x 1021 After 3 half lives no. of atoms = 1/8 x 3.01 x 1021 = 3.76 x 1020 • Calculations for you to try. • Th-234 has a half-life of 24.1 days. What mass of a 20.4g sample will remain after 96.4 days? At start mass = 20.4 g After 24.1 days mass left = 10.2 g After 48.2 days mass left = 5.1 g After 72.3 days mass left = 2.55 g After 96.4 days mass left = 1.275 g • A sample of Pu-242 has a mass of 1.21 g. • (a) How many atoms of Pu-242 are there in the sample? • (b) How many atoms of Pu-242 will remain after 3 half lives. • (c) Use the half life in the data book for Pu-242 to calculate the time it would take to reduce the number of Pu-242 atoms in the sample to 1/8 of its original value. (a) No of atoms = 1.21/242 x 6.02 x 1023. = 3.01 x 1021 • 1/8 of original value means that 3 half lives have passed. • 3 x 3.79 x 105 years = 1.137 x 106 years Higher Grade Chemistry
Calculations for you to try. • The count rate due to carbon-14 in ancient wooden timber was found to be • 100 counts per minute. A sample of modern wood had a count rate of 1600 counts per minute. Given that carbon-14 has a half life of 5570 years, calculate the age of the ancient timber. At start count rate = 1600 counts min-1. After 1 half-life = 800 counts min-1. After 2 half-life = 400 counts min-1. After 3 half-life = 200 counts min-1. After 4 half-life = 100 counts min-1. Age of timber = 4 x 5570 = 22 280 years. 4. A radioisotope used in a laboratory has a half life of 6.75 hours. It had a count rate of 2000 counts per minute at 8.00 a.m. on Monday. What would be the count rate at 11 a.m. the following day? Between 8.00 a.m. and 11.00 am the next day 27 hours have passed. Number of half-lives in 27 hours = 27/6,75 = 4 Higher Grade Chemistry Count rate 2000 1000 500 250 125 counts min-1.
