Question

$C_{6}H_{5}F^{18}$ is a $F^{18}$ radio-isotope labelled organic compound. $F^{18}$ decays by positron emission. the product resulting on decay is

• A. $C_{6}H_{5}O^{18} $
• B. $C_{6}H_{5}Ar^{18} $
• C. $B^{12}C_{6}H_{5}F$
• D. $C_{6}H_{5}O^{16} $

Answer 1

Answer: (A)

${ }_{9} F ^{18} \longrightarrow { }_{x} E ^{y}+{ }_{+1} e^{0}$
$x=8, y=18$
$\therefore { }_{x} E ^{y}={ }_{8} O ^{18}$
$\because$ Position has one unit of the charge and zero mass.
Thus, $F ^{18}$ changes to $O ^{18}$ therefore resulting decay product is $C _{6} H _{5} O ^{18}$.