A radioactive substance of half life 138.6 days is placed in a box. After (n ) days only (20 % ) of the substance is present then the value of (n ) is ([ (5)=1.61] )

Step-by-step solution

Half life of a radioactive substance, (n_ 1 / 2 =138.6 ) days If (N₀ ) be the initial amount of radioactive substance, then remaining amount after (n ) days is given by (N=20 % of N₀= 20 100 N₀= N₀ 5 ) By radioative decay's law, (N=N₀ ( 1 2 )^ n n_ 1 / 2 N₀ 5 =N₀ ( 1 2 )^ n 138.6 1 5 = ( 1 2 )^ n 138.6 ) Taking ( ) on the both sides, we get ( aligned 1 5 & = ( 1 2 )^ n 138.6 1 5 = n 138.6 ( 1 2 ) 5 & = n 138.6 2 n & =138.6 5 2 [ array l 5=1.61 2=0.693 array ] n & =138.6 1.61 0.693 n=322 days aligned )