Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 minutes.

In time t1, 33% decays , means 67% is  present no of nucleus=2N03  In time t2, 67% decays , means 33% is  present no of nucleus=N03 T1/2=20min ⇒ln⁡2λ=20min ⇒λ=ln⁡220(min)t=1λln⁡NoN ⇒t1=1λln⁡No2N03=1λln⁡32  &     t2=1λln⁡NoN03=1λln⁡3 Thus , t2−t1=1λln⁡3-ln⁡32=1λln⁡2=20min