How many disintegrations per second will occur in one gram of 92U238, if its half-life against α – decay is 1.42 x 1017 s ?

Q: How many disintegrations per second will occur in one gram of ₉₂U²³⁸, if its half-life against α – decay is 1.42 x 10¹⁷ s ?

Sol: Given Half- life period (T) = 0.693/λ = 1.42 × 10¹⁷ s

λ = 0.693/(1.42 × 10¹⁷ )= 4.88 × 10^-18

Avogadro number (N) = 6.023 × 10²³ atoms

n = Number of atoms present in 1 g of (₉₂U²³⁸) = N/A

= (0.623 × 10²³)/238 = 25.30 × 10^20

Number of disintegrations = dN/dt  = λ n

= 4.88 x 10^-18 × 25.30 × 10²⁰

= 1.2346 x 10⁴ disintegrates/sec.