The counting rate observed from a radioactive source at t=0 was 1600 counts s ⁻¹ , and t=8 ~s , it was 100 counts s ⁻¹ . The counting rate observed as counts s ⁻¹ at t=6 ~s will be

Step-by-step solution

As we know, aligned & [ N N₀ ]= [ 1 2 ]^n & n= no. of half life & N - no. of atoms left & N₀ - initial no. of atoms & By radioactive decay law, & d N d t =k N & k- disin t e g r a t i o n constant & d N d t d N₀ d t = N N₀ aligned From (i) and (ii) we get d N d t d N₀ d t = [ 1 2 ]^n or, [ 100 1600 ]= [ 1 2 ]^n [ 1 2 ]^4= [ 1 2 ]^n n=4 , Therefore, in 8 seconds 4 half life had occurred in which counting rate reduces to 100 counts s ⁻¹ . Half life, T₁ 2 =2 sec In 6 sec , 3 half life will occur aligned & [ d ~N d t 1