Question

What is the half‑life of an isotope that decays to 6.25% of its original activity in 70.6 h?

Answer 1

The decay of isotopes follow first order kinetics.

Now we have the relation

ln[A₀/A] = kt   …………(1)

where A₀ is the initial concentration of the isotope

A is the concentration of the isotope at time t

k rate constant

Here, A = 0.0625A₀ ,     t = 70.6 h

Using (1), ln[1/0.0625] = k x 70.6 => k = 0.03927 1/h

Now half life

t_1/2 = 0.693/k = 0.693/0.03927 h = 17.65 h

Hence half-life is 17.65 h.