If a substance is radioactive, this means that the nucleus is unstable and will

Question

If a substance is radioactive, this means that the nucleus is unstable and will therefore decay by any number of processes (alpha decay, beta decay, etc.). The decay of radioactive elements follows first-order kinetics. Therefore, the rate of decay can be described by the same integrated rate equations and half-life equations that are used to describe the rate of first-order chemical reactions. By manipulation of these equations, we can arrive at the following formula:

fraction remaining=AA0=(0.5)n

where A0 is the initial amount or activity, A is the amount or activity at time t, and n is the number of half-lives. The equation relating the number of half-lives to time t is

n=tt1/2

where t1/2 is the length of one half-life.

Part A
You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 29.1 minutes , what is the half-life of this substance?
Express your answer numerically in minutes.

Part B
An unknown radioactive substance has a half-life of 3.20 hours . If 24.0 g of the substance is currently present, what mass A0 was present 8.00 hours ago?
Express your answer numerically in grams.

Part C
Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 45.0 % of an Am-241 sample to decay?
Express your answer numerically in years.

Explanation / Answer

A)

we know that

for radioactive decay

A/Ao = (0.5)^n

so

100/400 = (0.5)^n

1/4 = (0.5)^n

n = 2

now

n = t / t1/2

so

2 = 29.1 / t1/2

t1/2 = 14.55 min

so

half life of the substance is 14.55 min

B)

now

n = t / t1/2

n = 8 / 3.2

n = 2.5

now

A/Ao = (0.5)^n

24 / Ao = (0.5)^2.5

Ao = 135.7645

so

135.7645 g of substance is present 8 hours ago

C)

given

45% is decayed

so

55% is remaining

so

A/Ao = 0.55

now

A/Ao = (0.5)^n

0.55 = (0.5)^n

log 0.55 = n x log 0.5

n = 0.8625

now

n = t / t1/2

so

0.8625 = t / 432

t = 372.6

so

the time taken is 372.6 years

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