Question

A sample of a radioactive substance decayed to 96.5% of its original amount after a year. (Round your answers to two decimal places.) (a) What is the half-life of the substance?

Answer 1

Answer:

The half-life of the substance is 19.47 years.

Step-by-step explanation:

The equation for the amount of substance remaining is given by the following equation:

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

In which Q(t) is the amount remaining after t years, Q(0) is the initial amount and r is the rate that this amount decreases.

A sample of a radioactive substance decayed to 96.5% of its original amount after a year.

This means that $Q(1) = 0.965Q(0)$

We use this to find r. So

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

$0.965Q(0) = Q(0)e^{-r]$

$e^{-r} = 0.965$

$\ln{e^{-r}} = \ln{0.965}$

$-r = =0.0356$

$r = 0.0356$

So

$Q(t) = Q(0)e^{-0.0356t}$

(a) What is the half-life of the substance?

This is t when Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0356t}$

$0.5Q(0) = Q(0)e^{-0.0356t]}$

$e^{-0.0356t} = 0.5$

$\ln{e^{-0.0356t}} = \ln{0.5}$

$-0.0356t = \ln{0.5}$

$t = -\frac{\ln{0.05}}{0.0356}$

$t = 19.47$

The half-life of the substance is 19.47 years.

Answer 2

Answer:

19.47 years

Step-by-step explanation:

In order to find the answer we need to use the following formula:

$m(t)=m(0)*e^{kt}$ where:

m(t)=amount after 't' years

m(0)=initial amount

k=decay constant

t= time in years

If the initial amount of substance is 'x' then the problem establishes:

m(0)=x*100

m(t)=x*96.5

t=1 year

Using the equation we have:

$x*96.5=x*100*e^{k*1}$

$ln(96.5/100)=k$

$-0.0356=k$

Now, the half-life of a substance is the time 't' which allows half of the substance to be decayed. So:

m(0)=x*100

m(t)=x*50

k=-0.0356

Then:

$x*50=x*100*e^{-0.0356t}$

$ln(50/100)/(-0.0356)=t$

$19.47=t$

In conclusion, the half-life of the substance is 19.47 years.