Question

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 28.1 minutes, what is the half-life of this substance?

Part B:

An unknown radioactive substance has a half-life of 3.20 hours. If 15.7 g of the substance is currently present, what mass A₀ was present 8.00 hours ago?

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?​

Part D:

A fossil was analyzed and determined to have a carbon-14 level that is 50 % that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?​

Answer 1

A)

we have:

[A]o = 400

[A] = 100

t = 28.1 min

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln(1*10^2) = ln(4*10^2) - k*28.1

4.605 = 5.991 - k*28.1

k*28.1 = 1.386

k = 4.933*10^-2 min-1

Given:

k = 4.933*10^-2 min-1

use relation between rate constant and half life of 1st order reaction

t1/2 = (ln 2) / k

= 0.693/(k)

= 0.693/(4.933*10^-2)

= 14.05 min

Answer: 14.0 min

B)

Given:

Half life = 3.2 hr

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(3.2)

= 0.2166 hr-1

we have:

[A] = 15.7 g

t = 8.0 hr

k = 0.2166 hr-1

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln(15.7) = ln[A]o - 0.2166*8

ln[A]o = 2.754 + 0.2166*8

ln[A]o = 4.486

[A]o = e^(4.486)

[A]o = 88.78 g

Answer: 88.8 g

C)

Given:

Half life = 4.32*10^2 years

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(4.32*10^2)

= 1.604*10^-3 years-1

we have:

[A]o = 100 (let initial amount be 100)

[A] = 55 (45 % has decayed. So, remaining is 55%)

k = 1.604*10^-3 years-1

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln(55) = ln(1*10^2) - 1.604*10^-3*t

4.007 = 4.605 - 1.604*10^-3*t

1.604*10^-3*t = 0.5978

t = 3.727*10^2 years

Answer: 373 years

D)

Since it contain half of initial concentration,

time taken = half life

= 5730 years

Answer: 5730 years