The decay constant of ²³⁸ U is 4.9 10⁻¹⁸ s ⁻¹ . By what factor will the activity of a ²³⁸ U sample decrease in 9 10⁹ years ?

Step-by-step solution

The half-life of ²³⁸ U is given by: T_ 1/2 = 2 = 0.693 4.9 10⁻¹⁸ s 1.414 10¹⁷ s Converting the half-life into years (using 1 year 3.14 10^7 s ): T_ 1/2 = 1.414 10¹⁷ 3.14 10^7 years 4.5 10^9 years The given time is t = 9 10^9 years . Number of half-lives, n = t T_ 1/2 = 9 10^9 4.5 10^9 = 2 The activity of the sample after n half-lives is A = A₀ 2^n . Substituting n = 2 : A = A₀ 2^2 = A₀ 4 Therefore, the activity decreases by a factor of 4 .