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)

$C_{6}H_{5}O^{18}$

Answer 2

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