1.

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 isA. 1000 sB. 1187 sC. 1200 sD. 1500 s

Answer» Correct Answer - b
`(N_(0))/(N)=(2.30)/(1.62)`, Now, `lambda = (2.303)/(600)log(2.30)/(1.62) = 0.000584`
`:. T_(1//2) = (0.693)/(lambda) = (0.693)/(0.000584) = 1187s`


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