`""^(18)F` is found to undergo 90% radioactive decay in 366 minutes, calculate half-life of `""^(18)F`.

Options

A. 90 minutes
B. 100 minutes
C. 105 minute
D. 110 minute

Answer

D

Explanation

`lambda = (2.303)/(t) log ""(N_(0))/(N)`
Here `lambda = (2.303)/(t) log "" (A_(0))/(A) = (2.303)/(366) = 6.29 xx 10^(-3)"minute"^(-1)`
`therefore t_(0.5) = (0.693)/(lambda) =(0.693)/(6.29 xx 10^(-3)) = 110` minute