Question

A 15-g sample of radioactive iodine decays in such a way that the mass remaining after t days is given by
m(t) = 15e−0.073t,
where
m(t)
is measured in grams. After how many days is there only 5 g remaining?

Answer 1

15 days

Step-by-step explanation:

The half-life equation is given as

$m(t) = m(0)e^{-0.073t}$ (1)

where m(0) is the initial mass of radioactive iodine at t = 0. To solve for the time t, let's put the exponential part on the left-hand side:

$e^{-0.073t} = \dfrac{m(t)}{m(0)}$

Taking the natural logarithm of both sides to get

$\ln{e^{-0.073t}} = \ln\left[\dfrac{m(t)}{m(0)}\right]$ (2)

Recall the following properties of logarithms:

$\ln{A}^b = b\ln{A}$

$\ln{e} = 1$

We can rewrite Eqn(2) as

$-0.073t = \ln\left[\dfrac{m(t)}{m(0)}\right]$

Solving for t, we get

$t = -\dfrac{1}{0.073}\ln\left[\dfrac{m(t)}{m(0)}\right]$

Since m(0) = 15 gm and m(t) = 5 gm, then the amount of time that elapses to reduce the mass down to 5 gm is

$t = -\dfrac{1}{0.073}\ln\left(\dfrac{5\:\text{g}}{15\:\text{g}}\right)$

$\:\:\:\:\:= 15\:\text{days}$