C_{6}H_{5}F^{18} is an F^{18} radio-isotope l | vedclass

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Question

(A) In positron emission,the atomic number of the nucleus decreases by $1$ while the mass number remains constant. The decay reaction for $F^{18}$ is:

Vedclass · Accepted Answer

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

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(A) In positron emission,the atomic number of the nucleus decreases by $1$ while the mass number remains constant. The decay reaction for $F^{18}$ is: ${ }_{9}F^{18} \longrightarrow { }_{8}O^{18} + { }_{+1}e^{0}$ Since the $F^{18}$ atom is part of the $C_{6}H_{5}F^{18}$ molecule,the $F^{18}$ nucleus transforms into an $O^{18}$ nucleus. Therefore,the resulting decay product is $C_{6}H_{5}O^{18}$.

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

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