Activity of a radioactive sample decreases to ( 1 4 )^ th of its original value in 4 days. Then in 16 days its activity will become x times the original value. The value of x is

Step-by-step solution

Let the initial activity be A₀ and the half-life be T_ 1/2 . The activity after time t is given by A = A₀ ( 1 2 )^ t T_ 1/2 . Given that A = A₀ 4 at t = 4 days: A₀ 4 = A₀ ( 1 2 )^ 4 T_ 1/2 ( 1 2 )^2 = ( 1 2 )^ 4 T_ 1/2 4 T_ 1/2 = 2 T_ 1/2 = 2 days. After t = 16 days, the activity A' is: A' = A₀ ( 1 2 )^ 16 2 = A₀ ( 1 2 )^8 = A₀ 256 Thus, the activity becomes 1 256 times the original value, so x = 1 256 . Answer: 1 256