Question

A nurse was accidentally exposed to Potassium-42 doing some brain scans for possible tumors. The error was not discovered until 356 hours later when the activity of the Potassium-42 was 2.0 microCi/g. If Potassium-42 has a half-life of 12 hours, what was the activity of the sample at the time the nurse was exposed?

Answer 1

Answer:

Explanation:

The activity of a radioactive sample is the number of decays that occurs per second and depends on the number of radioactive nuclei that remains. An old unit for activity that is still used is the curie. One curie is a large amount of radiation. This is because the curie was originally defined as the activity of one gram of radium. For this reason most working environments only use radioactive samples that have an activity on the order of microcuries

Answer 2

Given: Half life of Potassium-42: 12 hours

The activity of the Potassium-42 after 356 hours was 2.0 microCi/g.

To Find:  the activity of the sample at the time the nurse was exposed

Step by step Explanation:

Radioactive decay follows first order kinetics.

According to first order rate ;

$ln N= ln N_{0} - \beta t$

N= number of radioactive nuclei present after time t

$N_{0}$ is the initial number of nuclei present

$\beta$ is the decay constant

• For first order reaction, decay constant will be;

$\beta = \frac{0.693}{t_{1/2} }$

t1/2= half life of the nuclei

• On putting the values in the equation;

$ln N= ln N_{0} - \beta t\\\\ln N+ \beta t= ln N_{0} \\ln(2.0)= ln N_{0} - (\frac{0.693}{12} )\times 356\\0.693 + 20.559= ln N_{0}$

$21.252 = ln N_{0}$

• On taking antilog on both sides;

$N_{0}= 4.4095$

Hence, the initial concentration of potassium-42 to which nurse was accidently exposed is 4.4095 microCurie/g.