Question

A radioactive material is reduced to 1/16 of its original amount in 4 days. How much material should one begin with so that 4 × 10^–3 kg of the material is left after 6 days.

Answer 1

N//N0 = (1/2)ⁿ

where n = t/T is number of half lives. 

Half life, T = t/T = 4/4 = 1 day

If m₀ is initial mass of radioactive material, then m/m₀ =(1/2)ⁿ.

Here n= t/T = 6/1 = 6,m = 4 x 10^-3 kg

∴ m/m₀ =(1/2)⁶ = 1/64

or m₀ = 64 m = 64 × 4 × 10^–3 kg = 0.256 kg