Po-218 is an alpha emitter with a half-life of 3.0

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

ISBN: 9780321910295 34

Solution for problem 94P Chapter 17

Introductory Chemistry | 5th Edition

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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?

Step-by-Step Solution:

Step 1 of 3

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 $\times{}$ 1023 atoms

41.25 $\times{}$ 10-03 g X atoms

X...

Step 2 of 3

Chapter 17, Problem 94P is Solved

Step 3 of 3

Textbook: Introductory Chemistry

Edition: 5

Author: Nivaldo J Tro

ISBN: 9780321910295

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 2045 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 270 students have viewed the full step-by-step answer.