Question

A radioactive substance with a half-life of 3.0 days has an initial activity of 2400 Bq. What is its activity after 6.0 days?

Answer 1

Here,

half life , T = 3 days

at t = 6 days

for the initial activity , Ao = 2400 Bq

for the final activity

A = Ao * (0.50)^(t/T)

A = 2400 * (0.50)^(6/3)

A = 2400/4 = 600 Bq

the activity after 6 days is 600 Bq

Answer 2

SOLUTION 

Half life = 3 days and activity 2400 Bq

After 6 days activity = 2400* 1/2  in 3 days * 1/2  in next 3 days 

= 600 Bq (ANSWER)