Question

What is the half-life of polonium-210 if it takes 828 days for a sample to be reduced to 1.6% of its original mass?

Answer 1

The half-life of polonium-210 is approximately 138.792 days.

Step-by-step explanation:

We must remember that the decay of a radioisotope is modelled by this ordinary differential equation:

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

Where:

`m(t)` - Current mass of the isotope, measured in grams.

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

Whose solution is:

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

Where
`m_(o)` is the initial mass of the isotope, measured in grams.

Our first step is to determine the value of the time constant:

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

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

If we know that
`(m(t))/(m_(o)) = 0.016` and
`t = 828\,days`, then the time constant of the radioisotope is:

`\tau = -(828\,days)/(\ln 0.016)`

`\tau \approx 200.234\,days`

And lastly we find the half-life of polonium-210 (
`t_(1/2)`), measured in days, by using this expression:

`t_(1/2) = \tau \cdot \ln 2`

`t_(1/2) = (200.234\,days)\cdot \ln 2`

`t_(1/2)\approx 138.792\,days`

The half-life of polonium-210 is approximately 138.792 days.