Technetium-99m is an ideal radioisotope

Question

Technetium-99m is an ideal radioisotope for scanning organs because it has a half-life of 6.0 h and is a pure gamma emitter. Suppose that 79.5 mg  were prepared in the technetium generator this morning. How many milligrams of technetium-99  m would remain after the following intervals?

one half-life

Express your answer to three significant figures and include the appropriate units.

two half-lives

Express your answer to three significant figures and include the appropriate units.

18 h

24 h

Express your answer to three significant figures and include the appropriate units.

Answer 1

Anya

After every half life quantity reduces by a half

So after 1 half life, 79.5 x 0.5 = 39.75 mg left ( 39.8 to 3 significant figures)

after 2 half lives 79.5 x 0.5 x 0.5 = 19.875 = 19.9 g (3 significant figures)

and so on.

Note do not round intermediate values. For example for the 2 half lives don't calculate 79.5 x 0.5, giving 39.8 to 3sf, and then find 39.8 x 0.5, because that can give a small error in the final result

And then 18 h = 3 half lives, 24 h = 4 half lives etc. so you can use the same idea

Mike

Answer 2

Hey Anya,

Half-lives are the time it takes for half of a sample to radioactively decayed.

1. One half life later (6 hours) would mean that the amount of Technetium-99 left would halve once. 79.5mg/2= 39.8mg Technetium-99 remaining.
2. This would mean the initial amount of Technetium-99 is halved twice, so we can express this as 79.5mg/4, or 19.9mg Technetium-99 remaining.
3. 18 hours/6 hours/half life= 3 half lives. This means we halve the initial amount 3 times, or divide by 8 (1/2 *1/2*1/2). 79.5mg/8= 9.94mg Technetium-99 remaining
4. 24 hours/6 hours/half life= 4 half lives. This means we halve the initial 4 times (dividing by 16). 79.5mg/16= 4.97mg Technetium-99 remaining.

Hope this helps!