The half-life of a radioactive isotope is 3 h. if the initial mass of the isotope was 300 g, the mass which remained undercayed in 18 h would be

Options

A. 2.34 g
B. 1.17 g
C. 9.36 g
D. 4.68 g

Answer

D

Explanation

`t_(1//2)=3 hrs, t=18hrs`
No. of half lives, `n=(t)/(t_(1//2))=(18)/(3)=6`
`N=(N_(0))/(2^(n))=(300 g)/(2^(6))=4.6875 g`