If half life of an element is 69.3 hours then how much of its percent will decay between 10^ th and 11^ th hour. Initial activity =50 C _i .

Step-by-step solution

Let active Nuclei at t=0 is N₀ then, active nuclei at t=10 hour N₁=N₀ e^ -10 Active nuclei at t=11 hour N₂=N₀ e^ -11 aligned & % decay = ( N₁-N₂ N₁ ) 100 & = ( N₀ e^ -10 -N₀ e^ -11 N₁ ) 100 & = ( N₀ e^ -10 -N₀ e^ -11 N₀ e^ -10 ) 100 & = (1- 1 e^ ) 100= (1- 1 e^ 2 / T_ 1 / 2 ) 100 & = ( T_ 1 / 2 =69.3 hrs . ) & = (1- 1 e^ 0.01 ) 100 % decay =1 % . aligned