**Logarithms and Radioactive Decay**

Radioactive decay is another practical application of logarithms.

**About radioactive decay**

Radioactive substances decay by spontaneously emitting radiation. The rate of decay is proportional to mass. Physicists express decay rate in terms of half life. This is the length of time it takes a substance to decay so that half its mass is remaining, and therefore half the radioactive content remains.

Different radioactive materials have varying half life.

- Radioactive cesium, or cesium 137 has a half life of 30 years
- Radium, or radium-221, has a half life of 30 seconds
- The half life of radioactive carbon, known as carbon-14, is 5,730 years
- Uranium 234 has a half-life of 270,000 years
- The half life of krypton-91 is ten seconds

Typical questions:

- Compute the half-life of a material
- Compute the decay rate
- How much material is left after a certain amount of time
- How old is a sample
- How long until some percent of a sample remains
- What is the weight of the original sample

**One critical equation, and two more derived from it.**

The essential equation for radioactive decay questions using logarithms is:

m(t) = m(0) * e^(-rt)

m(0) is the original mass of the material

m(t) is the mass of the material following some time interval. This value is always some fractional portion of the original amount.

r is the decay rate

t is the length of time

Note the negative value of the exponent. As time progresses the exponential value will become greater, in the negative direction. An increasing negative exponent is the means the resulting expression will continue to get smaller and smaller. It will remain in this case a positive number but as the time (t) value gets larger the resulting expression continues to get smaller and approach zero.

From this equation answers to the previous questions can be computed with the use of logarithms.

The first and most important equations to derive are expressions for the half-life and decay rate.

It is important to note that the value of m(t) is some fractional percentage of m(0).

The first important calculation to make is to write an equation for half life. In this case m(t) is half of m(0), and time (t) is how long the half-life is.

1/2 m(0) = m * e^(-rt) divide both sides by m

1/2 = e^(-rt)

ln(1/2) = ln(e^(-rt)) take ln of both sides

ln(1/2) = -rt remember ln(1/2) is a negative number

r=ln(2)/t t is the half-life

r=ln(2)/h rate is the natural log of two over half-life

h=ln(2)/r half-life is natural log of two over rate

It is important to have an equation for half-life because it is frequently needed to solve questions about using logarithms to solve problems about radioactive decay

This leaves three key equations:

m(t) = m(0) * e^(-rt) … basic equation for computing amount remaining after a certain interval of time

1/2 m(0) = m(0) * e^(-rh) … application of the basic equation for expressing how much is left after one iteration of the half-life

r=ln(2)/h … based on the half-life equation the rate of decay can be expressed

It is important and relevant to remember that the log of 2 and 1/2 differ by the sign:

ln(1/2) = (-1) * ln(2)

**Example: Radioactive Decay:**

In this example the original amount of mass and rate of decay are given. m(0) is given as 100 grams and rate is 0.5 months.

m(t) = m(0) * e^(-0.5t)

When time (t) = zero then e^0=1 and m(0) = 100 grams. This means the original amount is 100 grams. The exponent is negative because the result, mass in grams, diminishes with time.

To find the amount two months later set t to 2 and solve the equation.

m(t) = 100 * e^(-0.5*2) = 100/e ~= 36.8 grams

To find the amount ten months later set t to 10 and solve the equation.

m(t) = 100 * e^(-0.5*10) = 100/e^5 ~= 0.67 grams

**Example:** Radioactive Decay, plutonium-241

In this example, plutonium-241 decays at a rate of 0.053 and the initial amount is 200 grams.

m(t) = m(0) * e^(-0.053t)

The objective is to compute the amount remaining after 0, 4, 10, and 20 years.

The rate of decay, initial quantity and equation for computing quantity remaining are given. To answer questions about how much material remains after some number of years, the time value, which is an exponent of e, is substituted and then the equation is solved.

At time = zero years, m(0) * e^(-0.053*0) = m(0) * 1 = m(0) = 200 grams

At time = 4 years, m(4) = m(0) * e^(-0.053*4) ~= 162 grams

At time = 10 years, m(10) = m(0) * e^(-0.053*10) ~= 118 grams

At time = 20 years, m(20) = m(0) * e^(-0.053*20) ~= 69 grams

**Example:** radioactive decay, polonium-210

Polonium-210 has a half-life of 140 days. Given a sample of 300 grams…

**Example:** radioactive decay, carbon-14

An Egyptian mummy cloth contains 59% of the carbon it contained originally. How old is the mummy cloth. The half-life of carbon-14 is 5,730 years.

Remember that m(t) = m(0) * e^(-rt)

Start by solving for rate (r) by computing half life. Doing this allows re-writing m(t) in terms of m(0) and this eliminates one of the variables from the equation. t in this case is 5,730.

1/2 * m(0) = m(0) * e^(-r*5730)

The only unknown in the equation is rate (r).

1/2 = e^(-r*5370) Divide both sides by m(0)

ln (1/2) = ln(e^(-r*5370)) Take ln of both sides

ln (1/2) = -r*5370 Simplify, ln(e)^x = x

-r = ln(1/2)/5370

r = ln(2)/5370 ln(1/2) = -ln(2)

Now that rate is computed plug in 0.59*m as m(t) and solve for t. If the carbon-14 residue is 59% of the original amount then m(t) = 0.59m(0). Time (t) is the only unknown in the equation:

0.59*m(0) = m(0) * e^(-(ln(2)/5730)*t) Divide both sides by m(0)

0.59 = e^(-(ln(2)/5730) * t) Take ln of both sides

ln(0.59) = ln(e^(-ln(2)/5730 * t) This makes the exponent go away

ln(0.59) = -ln(2)/5730*t Isolate t

ln(0.59)*5730/-ln(2) = t Simplify and remember ln(0.59)

is a negative number

t ~= 4,362

**Example:** radioactive decay, carbon-14

A bone sample contains 73 percent of the carbon-14 that a living organism would have. How long ago did the animal die?

The basic equation for solving radioactive decay problems is:

m(t) = m(0) * e^(-rt), where:

• m(t) is the mass after time t

• m(0) is the mass at time zero which is the starting time

• r is the rate of decay

• t is the amount of time that has elapsed

• and half life can be computed by setting m(t) = 1/2 m(0) and then solve for t.

• r can be computed once half-life is known

• In this example half life is already known and rate is 1/2 the natural log of two divided by half-life

Remember from the previous problem that, given a 5,730 year half-life, the rate of decay is ln(2)/5730

Using the known values the previous expression now is:

0.73*m(0) = m(0) * e^(-ln(2)/5370)*t) Objective is to solve for t

0.73 = e^(-ln(2)/5370)*t) Divide both sides by m(0)

ln(.73) = ln(e^(-ln(2)/5370)*t)) Take log of both sides

ln(.73) = -ln(2)/5370*t Isolate t

t = ln(.73)*5730/-ln(2) ln(.73) is a negative number

t ~= 2,602 years

**Example:** radioactive decay, radioactive iodine

Given a 15 gram sample of radioactive iodine with a decay rate of -0.087, after how many days is 5 grams remaining

In this case the only unknown is t:

m(t) is 5 grams

m(0) is 15 grams

r is 0.087

The expression is then: 5 = 15 e ^(-0.087 * t) and the objective is to solve for t. The first step would be to take the natural log of both sides. That way the exponent on e comes down:

5 = 15 * e^(-0.087*t) divide both sides by 15

1/3 = e^(-0.877*t) take natural log of both sides

ln (1/3) = ln(e^(-0.087 * t)) bring the exponent down

ln (1/3) = -0.087 * t isolate t

t = ln(1/3)/-0.087 remember ln(1/3) is a negative number

t ~= 12.6 days

**Example:** Radioactive Decay. Compute rate of decay

If the half-life of a substance is 140 days then compute the rate of decay.

To solve this problem start with the standard equation:

m(t) = m(0) * e^(-rt)

The half life is given. If time (t) is the half life then rate (r) can be solved for. At 140 days, m(t) is half the original amount m(0). In other words, m(t) = 1/2*m(0). Plugging in the known values gives an equation that looks like this:

1/2* m(0) = m(0) * e^(-r*140) divide both sides by m(0)

1/2 = e^(-r*140) take ln of both sides

ln (1/2) = ln(e^(-r*140)) bring down the exponent

ln (1/2) = -r * 140 isolate r

r = ln(1/2)/-140 ln of a number less than 1 is negative

r~= 0.00495

In other words, r = ln(2)/h ln(1/2) = -ln(2) and h = 140

**Example:** radioactive decay, radium-226.

Given a half-life of 1,600 years and a 22 mg sample, how much is left after 4,000 years and after how long will 18 mg remain.

The equation to work with is m(t) = m(0) * e^(-rt). It has been previously proven that rate of decay r = ln(2)/h where h is the half-life.

It is a good idea to approximate answers in order to help sanity-check calculations. 4,000 years is well over two half-lives so the result should be smaller than 1/2 * 1/2 * 22 = 5.5. If 18 mg is left that is significantly less than one half-life.

To answer the first question about how much is left after 4,000 years, set t = 4000, r=ln(2)/1600, and solve for m(t):

m(4000) = m(0) * e^(-rt)

m(4000) = 22 * e^(-ln(2)/1600) use a calculator

m(4000) ~= 3.9 mg

This result is reasonable compared to the approximation of 5.5.

To answer the second question of how much time will elapse to leave 18 mg remaining:

m(t), or amount at time t, is 18 mg

m(0), or amount at time 0, is 22 mg

rate r is ln(2)/h and h is 1600

The equation is m(t) = m(0) * e^(-ln(2)/1600*t) and the objective is to solve for time t:

18 = 22 * e^(-ln(2)/1600*t) divide both sides by 22. 18/22 = 9/11

9/11 = e^(-ln(2)/1600*t) take natural log of both sides

ln(9/11) = ln(e^(-1600*t)) bring down exponent

ln(9/11) = -1600*t isolate t

t = ln(9/11)/-1600 remember ln(9/11) is negative

t ~= 463 years. Considerably less than a half life this is reasonable.

Example, radioactive Cesium, Cesium-137

The half-life of cesium-137 is 30 years. Given a 10 gram sample.

• Compute the decay rate

• Determine how much remains after 80 years

• After how long will two grams remain

To compute the decay rate, set up the equation for a computation of half life.

The standard equation is m(t) = m(0) * e ^(-r*t)

For half life, m(t) is 1/2 m(0) and time (t) is 30 years

1/2 m(0) = m(0) * e^(-r*30) dividing both sides by m(0)

1/2 = e^(-r*30) take natural log of both sides

ln(1/2) = ln(e^(-r*30))

ln(1/2) = -r*30 remember ln(1/2) is negative

ln(2)/30 = r remember ln(1/2) = -ln(2)

r ~= 0.23

To compute how much mass is left after 80 years set up the standard equation for computing radioactive decay.

Estimate the residual amount in order to sanity check the computed results. Three cycles of the half life would be 90 years meaning about 10 grams * (1/2)^3 or about 1.25 grams. 80 years is less than 90 so the expected result is something close to and greater than 1.25 grams

The standard equation is:

m(t) = m(0) * e ^ (-r*t)

In this case m(t) is the unknown that the question must answer

m(0) is the original amount which is 10 grams

r is rate of decay which is ln(2)/h and the half life is given as 30 years

time t is given as 80 years

So the equation to solve is

m(80) = 10 * e^(-(ln(2)/30)*80

m(80) = 10e(-ln(2)*3/8)

m(80) ~= 1.57 grams

To compute how long until two grams remain, start with the same standard equation, substitute in the unknown values, and solve for time:

m(t) = m(0) * e^(-r*t)

m(t) is two grams

m(0) is the original 10 grams

rate r was previously computed as ln(2)/30

time (t) is what must be computed

2 = 10 * e ^(-ln(2)/30*t) divide both sides by ten to simplify

2/10 = e^(-ln(2)/30*t) take ln of both sides

ln(1/5)=e ^(-ln(2)/30*t)

ln(1/5) =-ln(2)/30*t ln of any number less than 1 is negative

ln(1/5) * 30 / -ln(2) = t both sides times 30/-ln(2)

t~=67.7 years

**Example:** Radioactive Strontium, strontium-90, has a 28 year half-life.

How long will it take 50 mg to decay to a mass of 32 mg. Note this is slightly more than half the original sample, so the computed time should be slightly less than the 28 year half-life.

Using the standard equation for radioactive decay:

m(t) = m(0) * e ^(-r*t)

m(t) is 32 mg

m(0) is 50 mg

rate r from previous computations is ln(2)/h. h is the half-life and is given as 28 years. So the full expression for rate r is ln(2)/28.

Substituting in the known values the expression for radioactive decay is

32mg = 50mg * e ^(-ln(2)/28*t) and the objective is to solve for time t

32/50 = e^(-ln(2)/28*t) divide both sides by 50 mg

ln(32/50) =e ^(-ln(2)/28*t) take natural log of both sides

ln(32/50)= -ln(2)/28*t remember ln of a fraction is a negative number

t = ln(32/50) * 28 / -ln(2) multiply both sides times 28/-ln(2) in order to isolate time t

t ~= 18 years

**Example:** Radioactive radium, radium-221, has a half-life of 30 seconds. How long will it take 95 percent of a sample to decay?

The process to solve this is to write out the standard equation for radioactive decay, substitute in the known values, and then solve for what is unknown. In this case the unknown value to solve for is time t.

The question is asking how long will it take for 95 percent of the sample to decay, which means in other words how long until only five percent is left.

m(t) = m(0) * e ^(-r*t)

m(t) is five percent of the original mass and is therefore equal to 0.05 * m(0)

m(0) is the original mass. A value is not given for this.

rate r is not given but was previously computed to be ln(2)/h where h is the half life which is given to be 30 seconds, so rate r = ln(2)/30.

time t is the time it takes for 95 percent of the sample to decay, meaning in other words, after what length of time is five percent of the original mass left?

One interesting thing in all this is that e works. It seems to be an interesting number

Substituting in the known values the expression for radioactive decay is

0.05(m0) = m(0) * e^(-ln(2)/30*t)

both sides can be divided by m(0). This eliminates one of the unknown values leaving time t as the only unknown.

(0.05) = e^(-ln(2)/30*t) take ln of both sides in order to bring down the exponent

ln(0.05) = ln(e^(-ln(2)/30*t))

ln(0.05) = -ln(2)/30*t to isolate t, divide both sides times 30/-ln(2)

t=ln(0.05) * 30 / -ln(2) remember ln(0.05) is a negative number

t ~= 130 seconds

This means that radioactive radium, also known as radium-221, with a half life of 30 seconds, will decay to five percent of its original mass in 130 seconds.

**Example: **If 250 mg of a radioactive element decays to 200 mg in 48 hours then what is the half life

From the basic equation for computing radioactive decay

m(t) = m(0) * e^(-rt), the available information is

m(t) is 200 mg

m(0) is 250 mg

time t is 48 hours

decay rate r is the unknown to be solved

In this case it is relevant to consider the expression for rate of decay. It has been previously established that the rate of decay is ln(2)/h, with h representing the half life.

Substituting in the known values the expression for radioactive decay is:

200 = 250 * (e^(-ln(2)/h*48)

This leaves only h as the unknown so this equation can be solved.

Dividing both sides by 250 gives

200/250 = e^(-ln(2)/h*48) 200/250 simplifies to 4/5

4/5 = e^(-ln(2)/30*t) take natural log of both sides

ln(4/5) = ln(e^(-ln(2)/h*48) simplify b/c ln(e^x) = x

ln(4/5) = -ln(2)/h*48 to isolate h multiply both sides by h/ln(4/5)

h = -ln(2)*48/ln(4/5) The negative signs cancel because ln(4/5) is a negative number

h ~= 149 hours

This means that if a 250 mg sample of a radioactive element decaus to 200 mg in 48 hours then the half-life of the element is 149 hours.

**Example:** A sample of radioactive radon, also called radon-222, decays to 58 percent of its original amount after three days. Compute the half-life and and time to decay to 20 percent of the original amount.

In this case it is relevant to consider the expression for rate of decay. It has been previously established that the rate of decay is ln(2)/h, with h representing the half life.

Starting with the standard expression for radioactive decay:

m(t) = m(0) * e^(-rt)

m(t) is equal to 58 percent of m(0)

time t is three days

rate r is ln(2)/h

So the expression with known values filled in is

0.58*m(0) = m(0) * e^(-r*3) dividing both sides by m(0) leaves rate r as the only unknown

0.58 = e^(-3r) Take ln of both sides to bring down the exponent

ln(0.58) = -3r Isolate r by dividing both sides by -3

r = ln(0.58)/-3 This result is a positive number because ln(0.58) is negative

It was previously proven that r = ln(2)/h so now there are two expressions for rate r with one unknown

ln(2)/h = ln(0.58)/-3 Isolate h by multiplying both sides by -3*h/ln(0.58)

h = ln(2)*-3/ln(0.58)

h ~= 3.82

The half life is 3.82 days. This would be consistent with slightly more residual (58 percent) in slightly less (3 days) than the half life.

To compute time to decay to 20% of the original amount, the place to start is with the equation for radioactive decay.

m(t) = m(0) * e^(-rt)

The objective is to isolate time t and compute its value. It is best to isolate the variable and then to plug in values.

First, divide both sides by m(0):

m(t)/m(0) = e^(-rt) next, take ln of both sides

ln (m(t)/m(0)) = ln(e^(-rt)) this allows removing the exponent

ln(m(t)/m(0)) = -rt divide by rate -r giving time t

t = ln(m(t)/m(0))/-r

In this case

m(t) = 20 percent of m(0) = 0.20(m0)

rate r has been previously calculated as ln(2)/h

h was previously calculated to be h = ln(2)*-3/ln(0.58)

substituting in all the known values creates a complicated looking expression that can carefully be simplified:

m(t)/m(0) = 0.2*m(0)/m(0) = 0.2

0.2 = e^(-rt)

ln(0.2) = ln(e^(-rt))

ln(0.2) = -rt remember r=ln(2)/h and h=ln(2)*(-3)/ln(0.58)

ln(0.2) = -r * t

t = ln(0.2)/-r

t=ln(0.2)/-(ln(.58)/-3) note three negative values in the denominator and one in the numerator meaning the expression is positive

t=ln(0.2)*(-3)/-ln(0.58)

t ~= 8.87 days

**Example: Carbon 14 dating.** A wood artifact contains 65 percent of the carbon 14 that is present in living trees. The question is how old is the piece of wood. The half-life of carbon-14 is 5,730 years.

The standard expression for radioactive decay is:

m(t) = m(0) * e^(-rt)

The objective is solve for t.

m(t) is equal to 65 percent of m(0) so the expression can be rewritten ias:

0.65*m(0) = m(0) * e^(-rt)

Both sides can be divided by m(0) which has the advantage of eliminating one of the unknown values.

0.65 = e^(-rt)

Taking the natural log of both sides will allow eliminating the exponential values

ln(0.65) = ln(e^(-rt))

ln(0.65) = -rt

t=ln(0.65)/-r Time t can now be isolated

rate of decay r has been previously computed as ln(2)/h and h for carbon-14 is 5,730 years, so rate r is ln(2)/5730. This can be substituted in to the equation.

t=ln(0.65)/-(ln(2)/5730)

t=ln(0.65)*5730/-ln(2)

ln(0.65) is a negative number so the expression is positive.

t ~= 3,561 years.

This answer is reasonable. More than half the carbon-14 is still in the sample so the age would be expected to be less than the half life of 5,730 years.

**Example: Carbon-14 dating.** A fabric has 59 percent of the carbon-14 that it contained originally. Given the half life of carbon-14 is 5,730 years, how old is the fabric?

The equation for radioactive decay is

m(t) = m(0)*e^(-rt)

The objective is to compute the value of time t so it will be isolated and then known values are substituted in.

Known values are:

m(t) = 59 percent of the original amount m(0) = 0.059 * m(0)

rate r is ln(2)/h where h is the half life, h = 5730, so r = ln(2)/5730

0.59 * m(0) = m(0) * e^(-rt) divide both sides by m(0)

0.59 = e^(-rt) take ln of both sides

ln (0.59) = ln(e^(-rt)) this allows the exponent to come down

ln(0.59) = -rt Substitute in the value of r

ln(0.59) = -ln(2)/5730 * t

t = ln(0.59)*5730/ln(1/2) remember -ln(2) = ln(1/2)

t ~= 4,362 years

The expected result would be some value less than the 5,730 year half-life of carbon-14 because just over 50 percent is left, and the computed result is within reason.

Example: Uranium-234. Half life is 2.7e5 or 270,000 years. Compute the amount of a 10mg sample remaining after 1,000 years, and how long it will take to decompose until 7 mg is left.

The basic expression for radioactive decay is

m(t) = m(0) * e^(-rt), where:

m(0) is 10 mb

rate r is ln(2)/h = ln(2)/270,000

time t is 1,000 years

The first objective is solve for m(t). Plugging in the known values results in the expression:

m(1000) = 10* e^(-ln(2)/270,000*1000)

m(1000) ~= 9.97 mg.

This answer is reasonable. 1,000 years is a tiny fraction of the half-life and only a small amount of the 10 mg sample would have decayed.

To answer the second question, about how long it will take to decompose the sample leaving 7mg, start again with the basic equation for radioactive decay

m(t) = m(0) * e^(-rt)

The objective is to solve for time t

The known values are:

m(t) = amount after time t = 7mg

m(0) = original amount of sample = 10 mg

half-life h is 270,000 years

decay rate r is ln(2)/h = ln(2)/270,000

The expression after substituting in the value of m(t) is

7 = 10 * e^(-rt)

7/10 = e^(-rt) take natural log of both sides

ln(7/10) = ln(e^(-rt)) this allows taking down the exponent

ln(7/10) = -rt

The next step is to substitute in the value of rate r

ln(7/10) = -ln(2)/270,000*t

The next step is to isolate time t by multiplying both sides by 270,000/ln(2)

t = ln(7/10)*270000/-ln(2) remember -ln(2) = ln(1/2)

t = ln(7/10)*270,000/ln(1/2)

t ~= 138,934 years

This would seem reasonable because residual amount 7 mg is greater than half the original amount so the expected result is some number of years less than the half-life.

**Example:** A sample of bismuth decayed to 33 percent of its original mass after 8 days. Compute the half life and mass remaining after 12 days.

Using the available information and the standard expression for radioactive decay, an expression for rate of decay can be computed and compared to the known computation of decay rate being equal to ln(2)/h

The standard expression for radioactive decay is:

m(t) = m(0) * e^(-rt)

The known information is:

m(t) is 33 percent of m(0)

time t is 8 days

The two unknown values are decay rate and original mass. Substituting in the known values

0.33*m(0) = m(0) * e^(-rt) dividing both sides by m(0) makes one of the two unknowns go away and results in an expression that can be solved:

0.33 = e^(-rt) &nbnbsp;

Taking the natural log of both sides makes the exponent go away.

ln(0.33) = ln(e^(-rt))

ln(0.33) = -rt isolating rate r gives

r = ln(0.33)/-t time in this case is 8 days

r = ln(0.33)/-8

Now there are two expressions for rate r, and one of the two expressions contains half-life:

ln(0.33)/-8 = ln(2)/h

h can be isolated by multiplying both sides of the expression by h*-8/ln(0.33)

h = -8 * ln(2) /ln(0.33)

h ~= 5 days

The half-life of bismuth-210 is 8 days. This answer is reasonable. In 8 days less more than half the mass disintegrates so the expected result would be less than 8 days.

The second question is how much mass remains after 12 days. To answer this question, like all the others, start first with the standard equation for radioactive decay:

m(t) = m(0) * e^(-rt)

The objective is to solve for m(t), which is mass remaining after 12 days.

Half-life has been computed and is about five days, from the expression -8*ln(2)/ln(0.33).

Time t is 12 days.

Rate r = ln(0.33)/-t = ln(0.33)/-8

Substituting in these values:

m(12) = m(0) *e^-(12r) divide both sides by m(0)

m(12)/m(0) = e^(-12r)

m(12)/m(0) = e^-(12*ln(0.33)/-8)

m(12)/m(0) ~= 0.1896 ~= 0.19 ~= 19 percent

After 12 days, nineteen percent, or about one fifth, of the sample is left.

**Example:** The half-life of radium-226 is 1,590 years.

Given a 150 mg sample write the equation to model mass remaining after t number of years

The basic equation for radioactive decay is

m(t) = m(0) * e^(-rt)

Rate r is equal to ln(2)/h, h is 1,590, so rate r is ln(2)/1590

This means 150 mg will have mass after t number of years is

m(t) = 150 * e^(-ln(2)/1590*t)

Compute mass remaining after 1,000 years

In this case time t = 1,000

m(1000) = 150 * e^(-ln(2)/1590*1000)

m(1000) ~= 97 mg

This answer would seem reasonable because 1000 years is greater than half the half life and the calculated residual amount is more than half the original amount

After how many years will 50 mg remain.

Half of 150 mg is 75, which is how much would remain after 1,590 years. A quarter of 150 mg is 37.5, which is how much would remain after 1,590 * 2 years, or two half-lives. This means the expected result will be less than two half-lives (1,590 * 2 years) and greater than one half-life of 1,590 years.

The basic equation for radioactive decay is

m(t) = m(0) * e^(-rt)

and the objective is to solve for time t

The known values are

m(t) is 50 grams

m(0) is 150 grams

half life h is 1,590 years

decay rate r is ln(2)/h = ln(2)/1,590

Time t is the unknown and the objective is to solve for time. The equation can be simplified if the mass values are substituted in first

50 = 150*e^(-rt) divide both sides by 150 and simplify the fraction

1/3 = e^(-rt) take ln of both sides to bring down the exponent

ln(1/3) = ln(e^(-rt))

ln(1/3) = -rt the equation will be a bit more simple to solve if the expression for rate is substituted in first before solving for time because this will eliminate the complication of a fraction over a fraction. rate r is ln(2)/h = ln(2)/1,590

ln(1/3) = -ln(2)/1,590*t

Multiply both sides by 1,590/ln(2) to isolate t

ln(1/3) * 1,590 / -ln(2) = t

Remember that ln(1/3) is negative because the natural log of any number less than one and greater than zero is a negative number. Since the fraction is a negative number divided by a negative number the result is positive. Working the equation on a calculator the result is

t ~= 2,520 years

This answer is reasonable given the expected result. It exceeds one half-life and is less than two half-lives.

**Example:** The half-life of palladium-100 is four days. After 20 days a sample mass is 0.375 grams.

Compute the initial mass.

To solve this problem start with the equation for radioactive decay and remember that decay rate is ln(2)/h

m(t) = m(0) * e^(-rt)

m(0) is the original mass and the objective is to compute this number

m(t) is given as 0.375 grams

half life h is given as four days

rate r is ln(2)/h so r = ln(2)/h

time t is given as 20 days

m(0) can easily be isolated in the original expression by dividing both sides by e^(-rt), giving:

m(0) = m(t)/e^(-rt)

substituting in the known values of r and t gives an exponent in the denominator that is a fraction and must be solved with care.

m(0) = 0.375/e^(-ln(2)/h*20))

m(0) ~= 12 grams

The original mass is 12 grams

Compute mass after 3 days

The basic equation for radioactive decay is

m(t) = m(0) * e^(-rt)

The objective is compute the value of m(t) when t=3

The known values are

m(0) is 12 grams as computed in the preceding step

half-life h is 4 days

rate of decay r is ln(2)/h, h=4, so r=ln(2)/4

Plugging the known values into the equation gives

m(3) = 12 * e^(-ln(2)/4*3) ~= 7.1 grams

The mass after three days is about 7.1 grams. This is a reasonable result. The time is just less than the half-life, and the residual quantity is marginally greater than half the original mass.

After how many days will 0.15 grams remain

Starting with the equation for radioactive decay:

m(t) = m(0) * e^(-rt)

m(t) is 0.15 grams

m(0) was previously computed and is 12 grams

half-life h was given as 4 days

decay rate r is ln(2)/h = ln(2)/4

The objective is solve for t based on the given values

To solve for t first divide both sides by m(0) and then take the natural log of both sides. Taking the natural log (ln) of both sides allows for eliminating the exponent so the following two expressions are equivalent.

ln(m(t)/m(0)) = ln(e^(-rt))

ln(m(t)/m(0)) = -rt

divide both sides by -r to isolate time t

t = (ln(m(t)/ln(m0))/-r

substitute in the value of r

t = (ln(m(t)/ln(m0))/(ln(2)/4)

the result is a complex looking fraction over a fraction that may have been more simple to solve by incrementally substituting in the known values

t = ln(0.15/12)/ln(2)/4

t ~= 25.3 days

This answer is reasonable given the half life of 4 days and a very small fraction of material remaining.

**Last example** The half-life of krypton-91 is ten seconds. If the initial sample is 3 grams, what is the equation for computing the amount remaining after t seconds.

This question is asking for a general expression rather than a specific answer. The task objective is to take the equation for radioactive decay and plug in the known values so that any value of time t can be used to determine a residual amount.

The equation to start with is

m(t) = m(0) * e^(-rt)

The known values are

half-life h is 10 seconds

rate of decay is ln(2)/h = ln(2)/10

initial mass m(0) is 3 grams

Substituting in the known values gives:

m(t) = 3 * e^(-ln(2)/10*t)

Using this expression the mass remaining after t number of seconds can be computed. For example, the amount remaining after 60 seconds is:

m(t) = 3 * e^(-ln(2)/10*60)

m(t) = 3 * e^(-ln(2)*6)

m(t) ~= 0.047 grams

How long will it take for the mass to decay to so the remaining mass is 10^-6 grams (0.000001 gram)

Starting with the expression for radioactive decay

m(t) = m(0) * e^(-rt)

m(0) is given as 3 grams

m(t) is the final quantity of 0.000001 grams

Half-life h is 10 seconds

rate r is ln(2)/h = ln(2)/10

The objective is to compute time t based on these values

0.000001 = 3 * e^(-ln(2)/10*t First, divide both sides by 3

There are several different usable notations here. 0.000001/3 = 1/3*10^6 = 1/3,000,000

0.000001/3 = e^(-ln(2)/10*t take ln of both sides to eliminate exponent

ln(1/3e6) = ln(e^(-ln(2)/10*t)

ln(1/3e6) = -ln(2)/10*t both sides times 10/-ln(2) to isolate t

ln(1/3e6)*10/-ln(2) = t

t ~= 215 seconds

In the last few cases values are substituted in for the variables one at a time rather than isolating the unknown and then plugging in all variables. This is because some of the resulting fractions are complex, such as a fraction over a fraction. It is easier to incrementally plug in the unknowns and easier to avoid mistakes by doing it this way.

**Summary**

pH is the negative base 10 log of the hydronium ion concentration in a fluid:

pH = -log[H3O]+

To isolate [H3O]+ exponentiate both sides and simplofy:

10^(pH) = 10^(-log[H3O]+)

10^(pH) = [H3O]+

The magnitude of an earthquake on a the Richter scale is the log of earthquake intensity divided by a constant representing a very minor earth quake. Intensity is an empirical measured number representing the amplitude of a wave form on a seismograph that recorded the earthquake. Magnitude is a computed number based on intensity.

M = log(I/S)

To compute the intensity, exponentiate both sides

10^M = 10^(Log(I/S)) = I/S

I = 10^M * S

The standard equation for computing radioactive decay is

m(t) = m(0) * e^(-rt) where

m(t) is the mass at a certain time interval

m(0) is the mass at time zero

r = rate of decay

t = time since time zero

decay rate can be computed if the half-life is known by substituting 1/2*m(0) for m(t) and then solving for time r. The result is that decay rate r is equal to ln(2)/h and half-life h is equal to ln(2)/r

Using m(t) = m(0) * e^(-rt) and r=ln(2)/h most questions about radioactive decay and carbon dating can be computed. Laws of exponentiating and taking the log of equations are used in these computations