A small amount of solution containing a radioactive nucleide A was administered into blood of a patient. The activity of the nucleide is `2 xx 10^(3)` dps. Its half life is 15 hours. After 5 hours a sample of the blood drawn out from the patient. It's activity was 16 dpm per mL.
Radioactive nuclide `A^(x)` decays by `beta^(-)` emission (42%) `beta^(+)` emission (58%) in the patient body Then half value of `beta^(-)` decay path is

Options

A. 27.2 hours
B. 10.2 hours
C. 3.57 hours
D. 35.72 hours.

Answer

D

Explanation

`42 = (lambda_(1))/((0.693)/(15)) xx 100 " " lambda_(1) = 0.0194 hr^(-1)`,
`t_(1//2)` of `beta` - emission process ` = 35.72` hrs.