The radioactive decay of Tl-206 to Pb-206 has a half-life of 4.20 min. Starting with 5.00 x 10^22 atoms of Tl, how many atoms remain after 42.0 min?

Question

The radioactive decay of ²⁰⁶Tl to ²⁰⁶Pb has a half-life of 4.20 min. Starting with 5.00 x 10²² atoms of Tl, how many atoms remain after 42.0 min?

Answer 1

Radioactive decay is a first order reaction, and as such will follow 1st order kinetics.

The integrated rate law for a first order reaction is

ln [A] = -kt + ln [A]_o and for a 1st order reaction t_1/2 = 0.693 / k

You can use this equation to solve this problem (see below for solution), but an easier method is to use

FR = 0.5ⁿ

FR = fraction remaining = ?

n = number of half lives that have elapsed = 42.0 min / 4.2 min = 10 half lives have elapsed after 42 min.

FR = 0.5¹⁰ = 0.0009766

Since this is the fraction remaining, we can now find the number of atoms remaining:

5.00x10²² atoms x 0.0009766 = 4.88x10¹⁹ atoms remain

(be sure to check the math)

Using ln [A] = -kt + ln [A]_o we have...

t_1/2 = 0.693/k

4.20 min = 0.693/k

k = 0.165

ln [A] = -(0.165)(42.0) + ln 5.00x10²²

ln [A] = -6.93 + 52.27

ln [A] = 45.31

[A] = 4.75x10¹⁹ atoms remain

(be sure to check the math)

The two values differ by only about 3%.