Respuesta :
Explanation:
(1) When a proton is converted into neutron then the positron emission takes place as follows.
[tex]^{1}_{1}p \rightarrow ^{1}_{0}n + ^{0}_{+1}e[/tex]
A positron is represented by the symbol [tex]^{0}_{+1}e[/tex]. Therefore, when a positron emission occurs then the resultant nuclei atomic number decreases by a unit mass.
General equation representing positron emission is as follows.
[tex]^{M}_{Z}A \rightarrow ^{M}_{Z-1}B + ^{0}_{+1}e[/tex]
Hence, Fluorine-18 decays by positron emission as follows.
[tex]^{18}_{9}F \rightarrow ^{18}_{8}O + ^{0}_{+1}e[/tex]
Therefore, when 18F undergoes positron emission, the product nucleus is, [tex]^{18}O[/tex].
(2) Expression for the half-life of a [tex]^{18}_{9}F[/tex] isotope is as follows.
[tex]t_{\frac{1}{2}} = \frac{0.693}{\lambda}[/tex]
[tex]\lambda = \frac{0.693}{1.83 hr}[/tex]
Initial activity, ([tex][N]_{o}[/tex]) = 4.00 millicuries
Activity after 95% = [tex]4.00 - 4.00 \times \frac{95}{100}[/tex]
= 0.20 millicuries
[tex]\lambda = \frac{2.303}{t} log \frac{[N]_{o}}{[N]_{t}}[/tex]
[tex]\frac{0.693}{1.83 hr} = \frac{2.303}{t} log \frac{4.00}{0.20}[/tex]
t = 7.912 hrs
Thus, we can conclude that it will take 7.91 hrs for 95% of the 18F to decay.
