Half-life of a radioactive substance is 30 days. Number of atoms in the substance is `10^12`. How many disintegration of atoms per second does occur?

Explanation

T=30 d = 30 x 24 x 60 x 60 s
`lambda=0.693/T`
`=(0.693)/(30xx24xx60xx60)`
`=2.67xx10^(-7)` (approx.)
If t=1s, `lambdat=lambda=2.67xx10^(-7)`
Hence, `e^(lambdat)` = 1.000000267 (approx.)
`N=N_0e^(-lambdat) = N_0/e^(lambdat) = 10^12/1.000000267`
`=9.99999733xx10^11` (approx.)
`therefore` Numbers of atoms disintegrated per second =`N_0-N = 2.67xx10^5` (approx.)