Question

Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of each element's decay is described by its half-life, which is the amount of time it takes for the number of radioactive atoms of this element to be reduced by half.
The half-life of the isotope beryllium-11 is 14 seconds. A sample of beryllium-11 was first measured to have 800 atoms. After t seconds, there were only 50 atoms of this isotope remaining.

Requried:
Write an equation in terms of t that models the situation.

Answer 1

Answer:

$Q(t) = 800e^{-0.0495t}$

Step-by-step explanation:

The amount of the isotope after t seconds is given by the following exponential equation:

$Q(t) = Q(0)e^{-rt}$

In which Q(0) is the initial amount and r is the decay rate.

A sample of beryllium-11 was first measured to have 800 atoms.

This means that $Q(0) = 800$

The half-life of the isotope beryllium-11 is 14 seconds.

This means that $Q(14) = 0.5Q(0)$

We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-14r}$

$e^{-14r} = 0.5$

$\ln{e^{-14r}} = \ln{0.5}$

$-14r = \ln{0.5}$

$r = -\frac{\ln{0.5}}{14}$

$r = 0.0495$

So

$Q(t) = Q(0)e^{-rt}$

$Q(t) = 800e^{-0.0495t}$