1.

The half-life of a radioactive isotope is 3 hour. IF the initial mass of isotope were `256g` the mass of it remaining undercayed after `18hr` is:A. `12g`B. `16g`C. `4g`D. `8g`

Answer» Correct Answer - `(c)`
`N_("Left") = (N_(0))/(2^(n)) = (256)/(2^(6)) = 4g (T = t_(1//2) xx n)`


Discussion

No Comment Found

Related InterviewSolutions