A sample of radioactive substance shows an intensity of 2.3 millicurie at a time t and an intensity of 1.62 millicurie after 600 s. The half-life period of the radioactive metal is

Explanation

`lambda = (2.303)/(t) log ((N_(0))/(N))`
`(0.693)/(t_(1//2)) = (2.303)/(600) log, ((2.3)/(1.62))`
`:. T_(1//2) = 1187` seconds