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

Here `(N^(0))/(N)= (2.30)/(1.62)`
We have, `lamda= (2.303)/(600) "log" (2.30)/(1.62)= 0.000584`
Now, `t_((1)/(2))= (0.6932)/(lamda)= (0.6932)/(0.000584)=1187` seconds