8 g of Cu 66 undergoes radioactive decay and after 15 minutes only 1 g remains. The half-life, in minutes, is then

Question

8 g of Cu 66 undergoes radioactive decay and after 15 minutes only 1 g remains. The half-life, in minutes, is then, 5, 15 / 8, 8 / 15, 15 ln ( 2 )

MARKS App · Accepted Answer

Given, m 0 = 8 g m = 1 g Time of decay = 15 min ln M M 0 = -λ t ⇒ ln 1 8 = -λ 15 ∴ λ = ln 8 15 So, T 1 / 2 = l n 2 λ = 15 ln 2 ln 8 = 5