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$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{F}^{18}$ is a $\mathrm{F}^{18}$ radio-isotope labelled organic compound. $\mathrm{F}^{18}$ decays by positron emission. The product resulting on decay is
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Verified Answer
The correct answer is:
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{18}$
${ }_{9} \mathrm{F}^{18} \longrightarrow{ }_{x} \mathrm{E}^{y}+{ }_{+1} e^{0}$
$$
x=8, y=18
$$
$\therefore \quad \mathrm{E}^{y}={ }_{8} \mathrm{O}^{18}$
$\because$ Position has one unit of the charge and
zero mass. Thus, $\mathrm{F}^{18}$ changes to $\mathrm{O}^{18}$ therefore resulting decay product is $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{18}$
$$
x=8, y=18
$$
$\therefore \quad \mathrm{E}^{y}={ }_{8} \mathrm{O}^{18}$
$\because$ Position has one unit of the charge and
zero mass. Thus, $\mathrm{F}^{18}$ changes to $\mathrm{O}^{18}$ therefore resulting decay product is $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{18}$
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