Question

The half-life of \[\ce{^238_92U}\] undergoing ∝- -decay is 4.5 × 10⁹ years. What is the activity of 1g sample of \[\ce{^238_92U}\]?

Answer 1

`T_(1/2) = 4.5 xx 10^9 y`

 = 4.5 × 10⁹ × 3.16 × 10⁷ s

= 1.42 × 10¹⁷ s

238g of isotope contains Avogadro's number of atoms, and so 1 gm \[\ce{^238_92U}\] Contains

`N = 1/238 xx 6.022 xx 10^23` atoms/kmol

∴ N = 25.3 × 10²⁰  atoms

The decay rate R is,

R = λN

= `0.693/T_(1/2)N`

= `(0.693 xx 25.3 xx 10^20)/(1.42 xx 10^17)s^-1`

= 1.23 × 10⁴ s^-1

=1.23 × 10⁴ Bq