Problem 94P
Po-218 is an alpha emitter with a half-life of 3.0 minutes. If a sample contains 55 mg of Po-218, how many alpha emissions occur in 6.0 minutes?
Solution:
Step 1
If the half life is 3.0 min, after 6.0 minutes the number of lifetimes will be
6.0 / 3.0= 2.
The final mass of Po will be:
M(Po-f) = Mo ( 1 / 2n = 55 )
1 / 22 = 55
55 / 4 = 13.75 mg = 0.01375 g
where n is the number of elapsed lifetimes.
If the original mass was 55 mg and the final mass is 13.75 mg, the amount of Po disintegrated will be 55-13.75 = 41.25 mg. The number of Po atoms in 41.25 mg may be calculated by direct proportionality:
218 g 6.02 1023 atoms
41.25 10-03 g X atoms
X = ( 41.25 10-3)
(6.02
1023) ] / (218) = 1.139
1020 atoms
1.14
1020 atoms
and this will also be the number of E(alpha-emissions), 1 alpha-particle / disintegrated atom.
Thus 1.14 1020 atoms of alpha emissions occur in 6.0 minutes
