Question

Bismuth-210 has a half-life of 5.0 days. (a) A sample originally has a mass of 800 mg. Find a formula for the mass remaining after t days. y(t) = 2 (b) Find the mass remaining after 40 days. mg (c) When is the mass reduced to 1 mg? (Round your answer to the nearest day.) days

Answer 1

a)
`y(t)=800e^(-0.1386t)`

b)
`3.1 mg`

c
`48.2 days`

Step-by-step explanation:

Let us first find the value of k, the rate constant;

k = (-ln0.5)/half life

= (-ln0.5)/5

= 0.1386

a)
`y(t) = y_(0)e^(-kt)\\y(t) = 800e^(-0.1386t)`

b)
`y(t) =800e^(-0.1386t) = 800e^(-0.1386*40) = 3.1 mg`

c
`y(t)=800e^(-0.1386t)\\1=800e^(-0.1386t)\\e^(-0.1386t)=1/800\\-0.1386tlne=ln(1/800)\\t=ln(1/800)/(-0.1386) = 48.2 days`

Answer 2

a) A(t) = 800 * e^-0.1386t

b) A = 3.1287 mg

c) t = 48.23 days

Step-by-step explanation:

First, this is a problem of exponential decay, so the general expression to use is the following:

A(t) = Ao*e^-kt (1)

Where:

Ao: Innitial mass, concentration, population.....

k: relative decay rate

t: time

A(t): mass, concentration or population left after t has passed.

Knowing this, the problem states that after 5 days, the bismuth decays to it's half life. So we can calculate the value of k, if we have a sample of 800 mg, after 5 days, we can say that this sample was reduced to 400 mg (Half of 800), therefore, we can replace the data on the above expression and find the formula for a mass remaining after t days:

A(t) = 400

Ao = 800

t = 5

k = ?

Replacing in equation (1) we have:

400 = 800 e^-5k ---> solving for k:

1/2 = e^-5k

ln(0.5) = -5k

k = -ln(0.5)/5

k = 0.1386

This is the value of the relative decay after t days. So if we replace in the original equation, we can find the formula for t days:

A(t) = 800 * e^-0.1386t

b) With the above formula, we only replace time and we'll get the remaining mass:

A = 800 * e^-0.1386(40)

A = 800 * 3.911x10^-3

A = 3.1287 mg

c) To know when the mass is reduced to 1 mg, we just solve for t from equation 1:

1 = 800 * e^-0.1386t

1/800 = e^-0.1386t

ln(1/800) = -0.1386t

t = -ln(1/800) / 0.1386

t = 48.23 days