The activity of a sample containing radioactive Ag is 6.4 x10 4th Bq. exactly 12 min later, its 2.0x10 3rd bq

Question

whats the half life the 4 and 3 should be superscripted

Answer 1

The activity of the sample is the number of disintegrations per unit time, which is given by the radioactive decay law

A = -dN/dt = λ·N

where lambda is the decay constant

Using the integrated form of the decay law, which is given by

N = N₀·exp(-λ·t)

(N₀ is the initial amount of radioactive atoms)

you find

A = λ·N₀·exp(-λ·t) = A₀·exp(-λ·t)

(where A₀ is the initial activity)

You know the activity at a given time, So solve this equation for the decay constant:

λ = - ln(A/A₀) / t = ln(A₀/A) / t

Decay constant and half life τ are related as

τ = ln(2) / λ

Therefore

τ = t · ln(2) / ln(A₀/A)

= 12min · ln(2) / ln(6.4×10⁴Bq / 2.0×10³Bq)

= 2.4min