Question

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?

Answer 1

Answer:

Time at which concentration is less than 25% = After 12:40 pm

Explanation:

Given:

• Half life of the reaction = 110 mins
• Time at which the injection was given = 9 am

To Find:

• The time at which the concentration becomes less than 25%

Solution:

Since this is an example of a radioactive decay, this is a first order reaction.

First finding the rate constant k,

The half life of a first order reaction is given by,

$\boxed{\tt t_{1/2}=\dfrac{0.693}{K}}$

Substitute the data,

$\tt 110\:min=\dfrac{0.693}{k}$

$\tt k=6.3\times 10^{-3}\:min^{-1}$

Now at the initial time t₀ let the concentration be R₀

At the final time t let the concentration be R

R = 25/100 × R₀

R = 0.25 R₀

Now the rate constant of a first order reaction is given by,

$\boxed{\tt k=\dfrac{2.303}{t} \: log\dfrac{[R_0]}{[R]}}$

Substituting the data,

$\tt 6.3\times 10^{-3}\:min^{-1}=\dfrac{2.303}{t} \: log\dfrac{R_0}{0.25\:R_0}$

$\tt 6.3\times 10^{-3}\:min^{-1}=\dfrac{2.303}{t} \: log\dfrac{1}{0.25}$

$\tt 6.3\times 10^{-3}\:min^{-1}=\dfrac{2.303}{t} \: log\dfrac{100}{25}$

$\tt 6.3\times 10^{-3}\:min^{-1}=\dfrac{2.303}{t} \: (log\:100-log\:25)$

$\tt 6.3\times 10^{-3}\:min^{-1}=\dfrac{2.303}{t} \: (2-1.3979)$

$\tt 6.3\times 10^{-3}\:min^{-1}=\dfrac{2.303}{t} \times 0.6021$

$\tt 6.3\times 10^{-3}\:min^{-1}=\dfrac{1.3866}{t}$

$\tt t=\dfrac{1.3866}{6.3\times 10^{-3}} \:min$

$\tt t=220.09\:min \approx 220\:min$

Hence after 220 mins the concentration of the isotope would be less than 25%.

Since the injection was given at 9 am, after 12:40 pm, the concentration would be less than 25%.