1) A doctor prescribes 200 milligrams of a drug that decays by about 14% each hour.

to the nearest minute, what is the half-life of the drug?

____ min

 

2) A wooden artifact contains 90% of the Carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half life of Carbon-14 is 5730 years,)

_____ Yr

Soln:- Here we have to solve the given problems by using exponential model and model of phenomenon of radio active decay.

Step-by-step explanation

Soln:- Here, amount of drug=x=200mg and rate of decay=14%=14/100=0.14

Therefore we can write it in exponential model

y=x(1-r)^t where x=initial value, r=rate and t= time

therefore by putting the values of x and r we get,

y=200(1-0.14)^t

y=200(0.86)^t --------------(1)

we have to find the half life of drug.

Therefore at half time , amount of drug left=200/2=100

Therefore, put y=100 in equation(1) we get,

100=200(0.86)^t

(0.86)^t =1/2

take log both side we get,

t*log(0.86)=log(1/2)

t=log(0.5)/log(0.86)

t=4.59 hr

t=4.59*60 min

t=276 min

The half-life of the drug is 276min

(2) soln:- Here we will use the formula Q(t) = Ae^rt to model the phenomenon of radio active decay.

where Q(t) is the amount of material after time t,

A is the the initial amount,

t is the time,

r is a constant depending on the material.

The half-life of carbon-14 is 5730 years, so

1/2 = e^5730r

take ln both sides we get,

ln(1/2)=5730r

r=0.000120968094 --------(1)

A wooden artifact contains 90 percent of the carbon-14 that is present in living trees, so here we will use the formula Q(t) = Ae^rt

0.9=e0.000120968094*t ------------(from eqn (1) we put value r)

take ln both sides we get,

ln(0.9)=0.000120968094*t

t=870.97 years

Approximately it is 870.97 years old