To go to the half-life calculator click here.

Actually, you don’t need to know about radioactive decay constants, λ , “k”, etc to do half-life calculations.
However, if you must learn about these in school, then this is the place to learn it.

If you need to know about the “k” radioactive decay constant, click here, otherwise just stay here.λ (lambda) is defined as the natural log of 2 divided by the half-life.

Plutonium 239 has a half-life of 24,100 years. What is lambda?

λ = ln (2) ÷ 24,100
λ = .693147 ÷ 24,100
λ = 2.876 × 10-5In this case the units for lambda would be years^-1 or 1/years.

A typical “half-life problem” might be worded:
Tungsten 181 has a λ value of 0.005723757/days.
What is its half-life?

Half-Life = ln(2) ÷ λ
Half-Life = .693147 ÷ 0.005723757
Half-Life = 121.1 days

Scroll down for 4 more half-life problems.

Here are the formulas used in calculations involving the exponential decay of radioactive materials.

Scroll down for 4 half-life problems.

1) You have 63 grams of cobalt 60 (half life = 5.27 years). How many grams will there be after 3 years?

λ = ln(2) ÷ half-life     λ = .69315 ÷ 5.27   λ = 0.13149 / year

Ending amount = beginning amount • e^{( -λ • time )}
Ending amount = 63 • 2.71828^{( -0.13149 • 3 )}
Ending amount = 63 • 2.71828^{( -.39448 )}
Ending amount = 63 • 0.67403
Ending amount = 42.464 grams

2) You measure 37 grams of strontium 90 (half life = 28.8 years). How much was present 2 years before this?

λ = ln(2) ÷ half-life     λ = .69315 ÷ 28.8   λ = 0.024068 / yearBeginning amount = ending amount ÷ e ^{( -λ • time )}
Beginning amount = 37 ÷ e^{( -.024068 • 2 )}
Beginning amount = 37 ÷ e^{( -0.048136 )}
Beginning amount = 37 ÷ .953
Beginning amount = 38.8 grams

3) You begin with 326.04 grams of iodine 131 and 11 days later, 126 grams remain.

What is the half life of iodine 131?

half life = [ time • ln (2) ] ÷ ln (beginning amount ÷ ending amount)
half life = [ 11 • .69315 ] ÷ ln (326.04 ÷ 126)
half life = [ 15.870 ] ÷ ln (2.5876)
half life = 7.6247 ÷ .95073
half life = 8.0198 days

4) Sodium 24 has a half life of 14.96 hours. How long will it take for a mass of sodium 24 to reach a mass that is just 5% of what you started with?

This might seem somewhat different than the other examples but all we have to do is let N_t = 5 and N₀ = 100.λ = ln(2) ÷ half-life     λ = .69315 ÷ 14.96   λ = 0.046333 / hour

time = ln (N_t ÷ N₀) ÷ -λ
time = ln (5 ÷ 100) ÷ -.046333
time = ln (.05) ÷ -.046333
time = -2.9957 ÷ -.046333
time = 64.66 hours

To go to the half-life calculator click here.

Actually, you don’t need to know about radioactive decay constants, λ , “k”, etc to do half-life calculations.
However, if you must learn about these in school, then this is the place to learn it.

If you need to know about the λ (lambda) radioactive decay constant, click here, otherwise just stay here.The radioactive decay constant “k” is defined as the natural log of 0.5 divided by the half-life.

Plutonium 239 has a half-life of 24,100 years. What is “k”?

k = ln (.5) ÷ 24,100
k = -.693147 ÷ 24,100
k = -2.876 × 10-5In this case the units for “k” would be years^-1 or 1/years.

A typical “half-life problem” might be worded:
Tungsten 181 has a “k” value of -0.005723757/days.
What is its half-life?

Half-Life = ln(.5) ÷ k
Half-Life = -.693147 ÷ -0.005723757
Half-Life = 121.1 days

Scroll down for 4 more half-life problems.

Here are the formulas used in calculations involving the exponential decay of radioactive materials.

1) You have 63 grams of cobalt 60 (half life = 5.27 years). How many grams will there be after 3 years?

k = ln(.5) ÷ half-life     k = -.693147 ÷ 5.27   k = -0.13149 / year

Ending amount = beginning amount • e^{( k • time )}
Ending amount = 63 • 2.71828^{( -0.13149 • 3 )}
Ending amount = 63 • 2.71828^{( -.39448 )}
Ending amount = 63 • 0.67403
Ending amount = 42.464 grams

2) You measure 37 grams of strontium 90 (half life = 28.8 years). How much was present 2 years before this?

k = ln(2) ÷ half-life     k = -.693147 ÷ 28.8   k = -0.024068 / yearBeginning amount = ending amount ÷ e ^{( k • time )}
Beginning amount = 37 ÷ e^{( -.024068 • 2 )}
Beginning amount = 37 ÷ e^{( -0.048136 )}
Beginning amount = 37 ÷ .953
Beginning amount = 38.8 grams

3) You begin with 326.04 grams of iodine 131 and 11 days later, 126 grams remain.

What is the half life of iodine 131?

half life = [ time • ln (2) ] ÷ ln (beginning amount ÷ ending amount)
half life = [ 11 • .69315 ] ÷ ln (326.04 ÷ 126)
half life = [ 15.870 ] ÷ ln (2.5876)
half life = 7.6247 ÷ .95073
half life = 8.0198 days

4) Sodium 24 has a half life of 14.96 hours. How long will it take for a mass of sodium 24 to reach a mass that is just 5% of what you started with?

This might seem somewhat different than the other examples but all we have to do is let N_t = 5 and N₀ = 100.

k = ln(2) ÷ half-life     k = -.693147 ÷ 14.96   k = -0.046333 / hour

time = ln (N_t ÷ N₀) ÷ k
time = ln (5 ÷ 100) ÷ -.046333
time = ln (.05) ÷ -.046333
time = -2.9957 ÷ -.046333
time = 64.66 hours