Half-life calculation using decay rates, please help!

Question

A radioactive isotope has an initial decay rate of 51.59 dps (disintegrations per second) for a sample. After an elapsed time of 99.48 days, the rate has decreased to 7.76 dps. What is the half-life (in days) of this isotope?

Answer 1

The change of the number of atoms ot the isotope with time is given by

N = N₀∙exp(- λ∙t)

with N₀ initial number of atoms and λ decay constant

Decay constant λ and half-life th are related as:

λ = ln(2)/th

The activity of the sample is given by

A = dN/dt = - λ∙N₀∙exp(- λ∙t) = A₀∙exp(- λ∙t)

Hence:

ln(A/A₀) = - λ∙t

<=>

ln(A/A₀) = - ln(2)∙t/th

<=>

ln(A₀/A) = ln(2)∙t/th

=>

th = ln(2) ∙ t / ln(A₀/A)

= ln(2) ∙ 99.48d / ln(51.59 / 7.76)

= 36.4 d