The radioactive isotope of lead, pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. if 1 gram of this isotope is present initially, how long will it take for 80% of the lead to decay? (round your answer to two decimal places.)

Apologies for the other person’s answer.

The half-life of an isotope describes the amount of time for half of the radioactive substance to decay into another form - in this case, the half life of this lead isotope is 3.3 hours.

We will use the two equations that are made for radioactive decay:

k = (ln(2))/ (t1/2)

k = (1/(to))ln(No/Nt),

where “t1/2” describes the half-life time (3.3 hours), “No” (actually “N zero”) refers to the amount of the original radioactive substance (how much was there initially), “Nt” refers to the amount of radioactive substance at some time “to”, and “to” (actually “t zero”) describes the amount of time required to reach the amount defined by “Nt”. It’s a lot.

We can assign the information given in the question to each of these:

t1/2= 3.3 hours
No = 1 gram
Nt = 0.2 grams (80% decayed)

Now, we just need to simplify the equations using these values.

k = ln2/(t1/2) = ln2 / 3.3 hours = 0.21

0.21 = (1/(to))ln(No/Nt) = (1/(to))ln(1/0.2) = (1/(to))(1.60944)

0.21 = (1.60944/(to))

to = 7.664 hours

It will take approximately 7.664 hours for the sample of lead to decay by 80 percent.

Hope this helps!

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