WBJEE · Chemistry · Chemical Kinetics
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{F}^{18}\) is a \(\mathrm{F}^{18}\) radio-isotope labelled organic compound. \(\mathrm{F}^{18}\) decays by positron emission. The product resulting on decay is
- A \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{18}\)
- B \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Ar}^{10}\)
- C \(\mathrm{B}^{12} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{F}\)
- D \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{16}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{18}\)
Step-by-step Solution
Detailed explanation
\({ }_{9} \mathrm{F}^{18} \longrightarrow{ }_{x} \mathrm{E}^{y}+{ }_{+1} e^{0}\) \[ x=8, y=18 \] \(\therefore \quad \mathrm{E}^{y}={ }_{8} \mathrm{O}^{18}\) \(\because\) Position has one unit of the charge and zero mass. Thus, \(\mathrm{F}^{18}\) changes to \(\mathrm{O}^{18}\)…
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