Question

based on the half-life of Tc-99, how many half-lives have to pass for a 150 mg sample of Tc-99 to decay down to 30mg?

Answer 1

2.322 half-lives have passed to decay down from 150 miligrams to 30 miligrams.

Step-by-step explanation:

The half of Technetium-99 is approximately 211000 years. The decay of isotopes is represented by the following ordinary differential equation:

`(dm)/(dt) = -(m)/(\tau)` (Eq. 1)

Where:

`(dm)/(dt)` - First derivative of isotope mass in time, measured in miligrams per year.

`m` - Mass of the isotope, measured in miligrams.

`\tau` - Time constant, measured in years.

Now we proceed to obtain the solution of this differential equation:

`(dm)/(m) = -(dt)/(\tau)`

`\int {(dm)/(m) } = -(1)/(\tau)\int dt`

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

`m(t) = m_(o)\cdot e^{-(t)/(\tau) }` (Eq. 2)

Where:

`m_(o)` - Initial mass of the isotope, measured in miligrams.

`t` - Time, measured in years.

The time passed for isotope is cleared within the equation described above:

`\ln (m)/(m_(o)) = -(t)/(\tau)`

`t = -\tau \cdot \ln (m)/(m_(o))`

In addition, we can obtain the time constant as a function of half-life:

`\tau = (t_(1/2))/(\ln 2)` (Eq. 3)

If we know that
`t_(1/2) = 211000\,yr`,
`m_(o) = 150\,mg` and
`m = 30\,mg`, then the time passed is:

`\tau = (211000\,yr)/(\ln 2)`

`\tau \approx 304408.654\,yr`

`t = -(304408.654\,yr)\cdot \ln \left((30\,mg)/(150\,mg) \right)`

`t \approx 489926.829\,yr`

The amount of passed half-lives is that time divided by a half-life. That is:

`n = (489926.829\,yr)/(211000\,yr)`

`n = 2.322`

2.322 half-lives have passed to decay down from 150 miligrams to 30 miligrams.