Half life of a radioactive substance is 20 minute. Difference between points of time when it is 33% disintegrated and 67% disintegrated is approximately

Given : T1/2=20 min. Decay constant, λ=0.693T1/2=0.69320=0.0345/ minute.we have N=N0e−λt Case-I: (When 33% disintegrated)N=N0−331000N0=671000N0 ∴671000N0=N0e−λt1                  ……..(1)∴671000e−λt1     or 0.67=e−λt1 t1=2.303λ(log0.67)=11.60 min.Case-II :(when 67% disintegrated)N2=N0−67100N0=33100N0 ∴33100=e−λt2                          ……….(2)or 0.33=e−λt2 or t2=2.303λlog(0.33)=32.14 min.Difference between time =t2−t1=20 min.