Question

A contaminated rag has been decaying for 48 hours and its activity measured to be 10.72% of its original activity. what it the half-life isotope?

A- 0.04652 hrsB- 48 hrsC- 24.09 hrsD- 14.9 hrs

Answer 1

14.9 hrs

Explanation: Let N₀ and N be numbers of atoms in the beginning and after 48 hours of duration respectively. λ be the disintegration constant.

`N = N_0e^{-\lambda 48`

dN₀ / dt = λN₀

dN / dt = λN

Given

dN / dt = .1072 x dN₀ / dt

λN = .1072 x λN₀

N = .1072 x N₀

N₀e^{-\lambda 48 =.1072 x N₀

e^{-\lambda 48 =.1072

Taking ln on both sides

- λ x 48 = ln .1072

λ x 48 = 2.233

λ = 2.233 / 48

`T_(1/2)` = .693 /λ

=( .693 /2.233 )X 48

= 14.9 hrs.