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,

$\frac{N}{N_{0}} = e^{- \lambda t}$

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

$\frac{N}{N_{0}} = e^{- \lambda t}$

Take natural logarithm "ln" on both sides

$ln [\frac{N}{N_{0}}]= - \lambda t$

Divide both sides by t

$\lambda = \frac{ln(N/N_{0})}{-t}$

Let us plug in the given values.

$\lambda = \frac{ln(8/64)}{-8}$

$\lambda = \frac{(-2.079)}{-8}$

$\lambda = 0.26$

The disintegration constant, λ is 0.26.

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

$t_{1/2} = \frac{0.693}{\lambda}$

Let us plug in the value of λ

$t_{1/2} = \frac{0.693}{0.26} = 2.7 hours$

The half life of the radioisotope is 2.7 hours.