Fluorine-18 (18f) is a radioactive isotope use in a variety of medical imaging procedures including positron emission tomography (pet) scans.
fluorine-18 decays by positron emission with a half-life of 1.83 hours.
(1) when 18f undergoes positron emission, the product nucleus is,
a) 18o b) 19ne c) 14n d) 17f
(2) a typical dose of 18f used for a pet scan has an activity of 4.00 millicuries. how long will it take for 95% of the 18f to decay?
a) 1.74 hrs b) 5.49 min c) 8.13 minutes d) 7.91 hrs.

Explanation:

(1)   When a proton is converted into neutron then the positron emission takes place as follows.

A positron is represented by the symbol . 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.

Hence, Fluorine-18 decays by positron emission as follows.

Therefore, when 18F undergoes positron emission, the product nucleus is,  .

(2)   Expression for the half-life of a isotope is as follows.

Initial activity, () = 4.00 millicuries

Activity after 95% =

                              = 0.20 millicuries

                          t = 7.912 hrs

Thus, we can conclude that it will take 7.91 hrs for 95% of the 18F to decay.

your answers are a,b,,e and f

i hope this

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