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?
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
Textbook: Introductory Chemistry
Author: Nivaldo J Tro
This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The answer to “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?” is broken down into a number of easy to follow steps, and 27 words. The full step-by-step solution to problem: 94P from chapter: 17 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This full solution covers the following key subjects: alpha, minutes, half, emitter, emissions. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. Since the solution to 94P from 17 chapter was answered, more than 335 students have viewed the full step-by-step answer.