**Problem #1:** Radioactive decay follows the following first-order law:

A = A_{o}e^{-kt}

where

A = activity at time t (sometimes you see it as A_{t})

A_{o}= initial activity (that is, when t = 0)

k = the decay constant

t = time

How can the above equation be rearranged to give (a) "k," (b) "t" and (c) "t_{1/2}" (the half-life)?

**Solution:**

1) Rearrange to get:

A/A_{o}= e^{-kt}

2) Take the natural log of both sides:

ln (A/A_{o}) = ln (e^{-kt})ln (A/A

_{o}) = -kt

3) This equation can be rewritten as:

ln A - ln A_{o}= -ktln A = -kt + ln A

_{o}and this last equation is commonly referred to as "the integrated form of the first-order rate law."

4) Now it is a simple matter to rearrange to solve for "k" or "t:"

(a) k = - (ln A - ln A_{o}) / t[another way: k = - ln (A/A

_{o})/t](b) t = - (ln A - ln A

_{o}) / k[another way: t = - ln (A/A

_{o})/k]

4) The solution for the half-life is a special case of (b) just above:
Set A_{o} equal to 2. At the end of one half-life, the activity, A, will be equal to 1 (which is one-half of 2). Thus, we have:

t_{1/2}= - ln (1/2) / kt

_{1/2}= (ln 2) / kThis last expression is commonly written as:

t

_{1/2}= 0.693 / k

Comment #1: the last expression can be rearranged to this:

k = 0.693 / t_{1/2}

Comment #2: the solution steps are: (1) determine the rate constant using the supplied data, then (2) use the above t_{1/2} equation to get the length of the half-life.

**Problem #2:** A 7.85 x 10^{-5} mol sample of copper-61 emits 1.47 x 10^{19} positrons in 90.0 minutes. What is the decay constant for copper-61?

**Solution:**

1) How many atoms in the sample before any decay?

7.85 x 10^{-5}mol times 6.022 x 10^{23}atoms/mol = 4.73 x 10^{19}atoms

2) What percent remains undecayed?

1.47 x 10^{19}/ 4.73 x 10^{19}= 0.311 (this is the amount that did decay.)1 - 0.311 = 0.689

3) Use integrated form of first-order rate law:

ln A = -kt + ln A_{o}ln 0.689 = - (k) (90.0 min) + ln 1

k = 0.00414 min

^{-1}

**Problem #3:** Plutonium-239 has a decay constant of 2.880 x 10^{-5} yr^{-1}. What percentage of a sample of Pu-239 remains after (a) 1400. years, (b) 2.100 x 10^{4} years, (c) 1.300 x 10^{5} years?

**Solution to a:**

ln A = -kt + ln A_{o}ln A = - (2.880 x 10

^{-5}yr^{-1}) (1400. yr) + ln 1ln A = -0.04032

A = 96.05% remaining

**Solution to b:**

ln A = -kt + ln A_{o}ln A = - (2.880 x 10

^{-5}yr^{-1}) (2.100 x 10^{4}yr) + ln 1ln A = -0.6048

A = 54.62% remaining

**Solution to c:**

2.366%

**Problem #4:** What is the decay constant of gallium-67, the half-life of which is 78.25 hr.

**Solution:**

k = (ln 2) / t_{1/2}k = (ln 2) / 78.25 hr

k = 8.858 x 10

^{-3}hr^{-1}

**Problem #5:** A certain radioactive isotope is considered to be safe when the concentration drops to 0.195% of its original amount. if the concentration of a sample drops from 0.50 M to 0.050 M in 1.00 years, how long (in years) will it take for the sample to be considered safe?

**Solution:**

1) The integrated form of the first-order rate law:

ln A = -kt + ln A_{o}

2) Calculate the decay constant:

ln 0.05 = - k (1.00 yr) + ln 0.5k = 2.3026 yr

^{-1}

3) Calculate the time to drop to 0.000975 M (which is is 0.00195 of 0.5 M):

ln 0.000975 = - (2.3026 yr^{-1}) t + ln 0.5t = 2.709 year (to two sig figs, this is 2.7 yr)

**Problem #6:** A radioactive isotope decays. If 17.0% of the isotope decays in 60.0 minutes, what is the half-life of the isotope?

**Solution:**

1) Since radioactive decay is first-order, we use the integrated form of the first-order rate law:

ln A = -kt + ln A_{o}

2) Calculate the decay constant:

0.17 gone means 0.83 remaining (it is necessary to use the amount remaining, not the amount decayed!)ln 0.83 = - k (60.0 min) + ln 1

-0.18633 = - k (60.0 min)

k = 0.0031055 min

^{-1}

3) Calculate length of half-life:

t_{1/2}= (ln 2) / kt

_{1/2}= 0.693 / 0.0031055 min^{-1}= 223 min

See here for a brief discussion (from the differential form of the first-order rate law) that shows the derivation of the t_{1/2} equation.

**Problem #7:** The radioisotope fluorine-21 had an initial mass of 80 milligrams. 20 milligrams of sample remained unchanged after 8.32 s. What is the half life of F-21?

**Solution:**

ln 20 / 80 = - k (8.32)- 1.39 = - k x 8.32

k = 0.167

t

_{1/2}= ln 2 / k = 0.693 / 0.167 = 4.16 s

This problem can also be solved as follows:

Take 80 and cut in half to get 40. That's one half-life.Now, take 40 and cut in half to get 20. A second half-life.

So, 8.32 sec extends over two half-lives.

8.32 / 2 = 4.16 sec

However, as you well know, not all problems are that easy. You have to know the general method using equations!

**Problem #8:** A radioactive isotope has a half-life of 1.2 billion years. As measured by the presence of the isotope and its stable decay product, a rock originally contained 10 grams of the radioactive isotope, and now contains 1.25 grams. Approximately how many years old is the rock?

**Solution:**

A small variation used on the notation used in probem #1 (N is used in place of A):

N_{t}= N_{o}times (1/2)^{(-t/τ)}N

_{t}= amount of matter at given time

N_{o}= original amount

τ (lowercase Greek letter tau) = half-life, also written as t_{1/2}1.25 = 10 times (1/2)

^{( -t / (1.2 x 109) )}t = ln(1.25/10)/ln(2) times -1.2 x 10

^{9}t = 3.6 x 10

^{9}yr

The solution using the formula developed in problem #1. First, the decay constant:

k = (ln 2) / t_{1/2}k = (ln 2) / 1.2 x 10

^{9}k = 5.776 x 10

^{-10}yr^{-1}Then:

ln 1.25 = - (5.776 x 10

^{-10}yr^{-1}) (t) + ln 10ln 0.125 = - (5.776 x 10

^{-10}yr^{-1}) (t)t = 3.6 x 10

^{9}yr

This answer technique:

10 divided by 2, divided by 2, divided by 2 = 1.25 g = 3 half lives3 x 1.2 billion = 3.6 billion years

is only available because of a round number of half-lives was involved.

**Problem #9:** The radioisotope potassium-40 decays to argon-40 by positron emission with a half life of 1.3 x 10^{9} years. A sample of rock was found to contain 78 argon-40 atoms for every 22 potassium-40 atoms. What is the age of the rock?

**Solution:**

1) Decimal amount of K-40 remaining:

22 / (78 + 22) = 0.222) Decay constant:original amount of K-40 = 1.00 (the decimal equivalent of 100%)

k = (ln 2) / 1.3 x 10^{9}yr = 5.3319 x 10^{10}yr^{1}

2) First-order rate law:

ln A = -kt + ln A_{o}ln 0.22 = - (5.3319 x 10

^{10}yr^{1}) (t) + ln 1.00t = 2.84 x 10

^{9}yr

**Problem #10:** The half life of strontium-90 is 28.0 years. Deduce that the decay constant of strontium-90 is 7.84 x 10^{-10} s^{-1}

**Solution #1:**

28.0 years times (365.25 day/yr) times (24 hr / day) times (3600 s / hr) = 883612800 sk = (ln 2) / t

_{1/2}k = (ln 2) / 883612800 s

k = 7.8444674 x 10

^{-10}s^{-1}rounded off to three sig figs, this is 7.84 x 10

^{-10}s^{-1}

**Solution #2:**

ln A = -kt + ln A_{o}ln 0.5 = - (k) (883612800 s) + ln 1

k = 7.84 x 10

^{-10}s^{-1}

Comment: in one half-life, the amount of Sr-90 is cut in half, thus A = 1, being cut in half from its starting value of A_{o} = 2 in one half-life.

**Bonus Problem:** If 1.000 x 10^{-12} mol of Cs-135 emits 1.390 x 10^{5} beta-particles in 1.00 year, what is the decay constant?

**Solution:**

1) What decimal percent remains after 1 year of decay?

1.00 x 10^{-12}mol times 6.022 x 10^{23}atoms/mole = 6.022 x 10^{11}atoms (this is A_{o})6.022 x 10

^{11}atoms minus 1.390 x 10^{5}= 6.02199861 x 10^{11}atoms6.02199861 / 6.022 = 0.99999977

2) Use integrated form of first-order rate law:

ln A = -kt + ln A_{o}ln 0.99999977 = - (k) (1.00 yr) + ln 1

k = 2.3 x 10

^{-7}yr^{1}

Comment: the same answer would result from this:

ln 6.02199861 = - (k) (1.00 yr) + ln 6.0221.79541919923919 = - (k) + 1.795419430059542

k = 0.000000230820352 = 2.3 x 10

^{-7}yr^{1}