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. $C_6{H}_{5}A{r}^{19}$
• C. $B^{12}{C}_5{H}_{5}F$
• D. $C_6{H}_5{O}^{16}$

Answer 1

Answer: (A)

$C_6{H}_5{O}^{18}$

The positron has one unit of the charge and zero mass.

${}_9{F}^{18}\to {}_x{E}^y+{}_{+1}{e}^0$

$x=8,y=18,$

$\therefore {}_x{E}^y={}_8{O}^{18}$

$C_6{H}_5{F}^{18}\to$ $C_6{H}_5{O}^{18}$
Due to positron emission, ${}_9{F}^{18}$ is converted to ${}_8{O}^{18}$.

${}_9{F}^{18}{\to }_8{O}^{18}{+}_{+1}{\beta }^0$

Hence, option A is correct.