Question

Carbon 14 is a radioactive isotope of carbon, the most common isotope of carbon being carbon 12. Carbon 14 is created when cosmic ray bombardment changes nitrogen 14 to carbon 14 in the upper atmosphere. The resulting carbon 14 combines with atmospheric oxygen to form radioactive carbon dioxide, which is incorporated into plants by photosynthesis. Animals acquire carbon 14 by eating plants. When an animal or plant dies, it ceases to take on carbon 14, and the amount of isotope in the organism begins to decay into the more common carbon 12. Carbon 14 has a very long half-life, about 5730 years. That is, given a sample of carbon 14, it will take 5730 years for half of the sample to decay to carbon 12. The long half-life is what makes carbon 14 dating very useful in dating objects from antiquity.

Consider a sample of material that contains A(t) atoms of carbon 14 at time t. During each unit of time a constant fraction of the radioactive atoms will spontaneously decay into another element or a different isotope of the same element. Thus, the sample behaves like a population with a constant death rate and a zero birth rate. Make use of the model of exponential growth to construct a differential equation that models radioactive decay for carbon 14.

Solve the equation that you proposed in to find an explicit formula for A(t).

Answer 1

Differential equation:
`(dA)/(dt) = k\cdot A(t)`,
`k < 0`,
`k = -(1)/(\tau)`,
`\tau > 0`.

Solution:
`A(t) = A_(o)\cdot e^{-(t)/(\tau) }`

Explanation:

Let suppose that rate of change of the amount of Carbon-14 atoms are directly proportional to the current amount of Carbon-14 atoms. That is:

`(dA)/(dt)\propto A(t)`

`(dA)/(dt) = k\cdot A(t)`

Where:

`A (t)` - Amount of Carbon-14 atoms, dimensionless.

`k` - Proportionality constant, measured in
`(1)/(yr)`.

`k` must be negative as death rate is constant and birth rate is zero. (
`k< 0`). Dimensionally, we can rewritte this constant as following:

`k = -(1)/(\tau)`

Where
`\tau` is the time constant (
`\tau > 0`), measured in years.

We can find the solution of the ordinary differential equation by separating each variable:

`(dA)/(dt) = -(1)/(\tau)\cdot A`

`\int {(dA)/(A) } = -(1)/(\tau) \int \, dt`

`\ln A(t) = -(t)/(\tau) + C`

`A(t) = e^{-(t)/(\tau)+C }`

`A(t) = e^(C)\cdot e^{-(t)/(\tau) }`

`A(t) = A_(o)\cdot e^{-(t)/(\tau) }`

Where
`A_(o)` is the initial amount of atoms of Carbon-14.