Question

^85 Sr is a short-lived (half-life 65 days) isotope used in bone scans. A typical patient receives a dose of ^85 Sr with an activity of 0.10 mCi. If all of the ^85 Sr is retained by the body, what will be its activity in the patient's body after one year has passed? answer in muCi

Answer 1

After one year has passed, the activity of ^85 Sr in the patient's body will be approximately 0.00014 muCi.

Step-by-step explanation:

^85 Sr is a short-lived isotope with a half-life of 65 days. To calculate the activity after one year, we can use the decay equation:

`\[ \text{Final Activity} = \text{Initial Activity} * \left( (1)/(2) \right)^{\frac{\text{time elapsed}}{\text{half-life}}} \]`

The initial activity is given as 0.10 mCi, and the time elapsed is one year, which is equivalent to 365 days. Substituting these values into the equation:

`\[ \text{Final Activity} = 0.10 \, \text{mCi} * \left( (1)/(2) \right)^{(365)/(65)} \]`

Solving this expression yields the final activity in mCi, which can be converted to muCi by multiplying by 1000. The result is approximately 0.00014 muCi.

In the calculation, the exponent represents the number of half-lives that have passed. In one year, there are about 365/65 ≈ 5.615 half-lives. Each half-life reduces the activity by half. Therefore, raising 1/2 to the power of 5.615 reflects the cumulative decay over the year.

It's important to note that the activity is decreasing over time due to the radioactive decay of ^85 Sr. This calculation assumes that all the administered ^85 Sr is retained by the body and that no additional sources of the isotope are introduced during the one-year period.

Answer 2

The remaining activity of the isotope ^85Sr in the patient's body after one year is approximately 1.42 μCi, computed using the half-life formula.

Step-by-step explanation:

To determine the remaining activity of the isotope ^85Sr in the patient's body after one year, we can use the concept of half-life. The half-life of ^85 Sr is 65 days, which means that every 65 days, the activity will reduce to half of its initial value.

The number of half-lives that have passed in one year (365 days) can be calculated as:

Number of half-lives = Total time elapsed / Half-life period

Number of half-lives = 365 days / 65 days = 5.615 (approximately)

To find the final activity after one year, we use the following formula:

Final activity = Initial activity x (1/2)^(Number of half-lives)

Plugging in the values:

Final activity = 0.10 mCi x (1/2)
`^{5.615`

This results in a final activity of approximately 0.00142 mCi, which is equivalent to 1.42 μCi.