Chemistry, asked by Vikas3075, 5 months ago

Fluorine-18 is a radioactive isotope of fluorine that is used in medical imaging scans. It has a
half-life of approximately 110 minutes. If a patient has a medical imaging scan using an
injection of fluorine-18 at 9 am, at what time will there be less than 25% of the radioactive
isotope in her body?
a) 10 am
b) 12 pm
c) 1 pm

Answers

Answered by abhi178
5

Given info : Fluorine-18 is a radioactive isotope of fluorine that is used in medical imaging scans. It has a half-life of approximately 110 minutes. If a patient has a medical imaging scan using an

injection of fluorine-18 at 9 am

To find : at what time will there be less than 25% of the radioactive isotope in her body ?

solution : half life of radioactive decay, T½ = ln2/λ

⇒λ = ln2/110 min¯¹

now initial amount = N₀

final remaining amount, N = 0.25N₀

using formula, t = 1/λ ln[N₀/N]

= 1/(ln2/110) × ln(1/0.25)

= 110/ln2 × ln4

= 220 minutes

= 3 hrs 40 minutes

so, definitely time taken to remain less than 25 % is more than 3 hrs 40 minutes.

only option (4) is satisfied above statement so option (4) is correct choice

Answered by Ranveerx107
12

half life of radioactive decay, T½ = ln2/λ

⇒λ = ln2/110 min¯¹

now initial amount = N₀

final remaining amount, N = 0.25N₀

using formula, t = 1/λ ln[N₀/N]

= 1/(ln2/110) × ln(1/0.25)

= 110/ln2 × ln4

= 220 minutes

= 3 hrs 40 minutes

so, definitely time taken to remain less than 25 % is more than 3 hrs 40 minutes.

only option (4) is satisfied above statement so option (4) is correct choice

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