Question

Uranium disintegrates at a rate proportional to the amount present at any time instant. If a and b grams of uranium are present at times t’ & t’’ respectively. What is the expression for the half life of uranium?

(a) \(\frac{(t”+t’)log2}{log⁡(\frac{b}{a})} \)

(b) \(\frac{(t”-2t’)log2}{2log(\frac{a}{b})} \)

(c) \(\frac{(t”+t’)}{2log(\frac{a}{b})} \)

(d) \(\frac{(t”-t’)log2}{log⁡(\frac{b}{a})} \)

This question was addressed to me by my school teacher while I was bunking the class.

This intriguing question originated from Law of Natural Growth and Decay topic in division Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct choice is (d) \(\frac{(t”-t’)log2}{log⁡(\frac{b}{a})} \)

Best explanation: Let P grams of uranium be present at any time t. According to law of decay, \(\frac{dp}{dt} = -kP…k\) is a constant its solution is P=ce^-kt given that at t=t’ P=a

a=ce^-kt^‘…..(1) and b=ce^-kt^‘….(2) dividing (1) & (2) we get \(\frac{a}{b}=e^{k(t^{”}-t’)}\)

taking log on both side \(log(\frac{a}{b})\) = k(t’’-t’)…(3) at t=0 ‘c’ is the initial mass of uranium present in the sample thus at half life  i.e t=T ,P=c/2 P = P=ce^-kt becomes \(\frac{c}{2}\) = ce^-kT

log 2=kT substituting the value of k from (3) we get \(T = \frac{(t”-t’)log2}{log⁡(\frac{b}{a})} \) is the equation for half life period.