Over a duration of 20 hours , the count rate from a radioactive sample drops from 4.0 10⁶ per second to 1.0 10⁶ per second. Calculate the count rate 100 hours after the initial instant.

Step-by-step solution

The initial count rate is R₀ = 4.0 10⁶ s ⁻¹ . After t = 20 hours , the count rate is R = 1.0 10⁶ s ⁻¹ . Using the law of radioactive decay, R = R₀ ( 1 2 )^ n , where n is the number of half-lives. 1.0 10⁶ = 4.0 10⁶ ( 1 2 )^ n ( 1 2 )^ n = 1 4 = ( 1 2 )² n = 2 Since 2 half-lives correspond to 20 hours , the half-life is T_ 1/2 = 10 hours . For t' = 100 hours , the number of half-lives is n' = 100 10 = 10 . The count rate after 100 hours is R' = R₀ ( 1 2 )¹⁰ . R' = 4.0 10⁶ 1 1024 R' = 4000 10³ 1024 3.9 10³ per second