A radioactive isotope decays to such a rate that after 96 min only 1/8th of the original amount remains. The value `t_(1//2)` of this nuclide is

Options

A. 12
B. 24
C. 32
D. 48

Answer

C

Explanation

`lamda = (2.303)/(t) "log" ([N_(0)])/([N]) = (2.303)/(96) "log" (1)/(1//8)`
`:. T_(1//2) = (0.693)/(lamda) = (0.693)/(0.0216) = 32.0` min