Question

A radioisotope is placed near a radiation detector, which registers 64 counts per second. Eight hours later, the detector registers 8 counts per second.What is the half-life of the radioactive isotope?

Answer 1

Radioactive decay follows first order kinetics

ln (No/ Nt) = kt

k = rate constant

t = time

N0 = initial count = 64

Nt = count at time ti = 8

So

K = ln (64/8) / time = 2.0794 /8 = 0.2599 hours-1

half life = 0.693 / K = 0.693 / 2.599 = 2.67 hours

Answer 2

Answer : The half life of the radioisotope is 2.7 hours.

Explanation :

Radioactive disintegration is a first order reaction. The disintegration equation can be written as,

`(N)/(N_(0)) = e^(- \lambda t)`

Here N is the amount of radioactive substance left = 8 counts

No is the initial amount of Radioactive substance = 64 counts

t is the time = 8 hours

λ is the disintegration constant

Let us rearrange the equation to solve for λ

`(N)/(N_(0)) = e^(- \lambda t)`

Take natural logarithm "ln" on both sides

`ln [(N)/(N_(0))]= - \lambda t`

Divide both sides by t

`\lambda = (ln(N/N_(0)))/(-t)`

Let us plug in the given values.

`\lambda = (ln(8/64))/(-8)`

`\lambda = ((-2.079))/(-8)`

`\lambda = 0.26`

The disintegration constant, λ is 0.26.

λ and half life ( t1/2) are related to each other by following equation.

`t_(1/2) = (0.693)/(\lambda)`

Let us plug in the value of λ

`t_(1/2) = (0.693)/(0.26) = 2.7 hours`

The half life of the radioisotope is 2.7 hours.