Phosphorus- 32 is commonly used in nuclear medicine for the identification of malignant tumors. It decays to sulphur- 32 with a half-life of 14.29 days. If a patient is given 3.5 mg of phosphorus-32, how much phosphorus-32 will remain after 1 month (i.e. 30 days)?

Short Answer

Expert verified
After 1 month (30 days), approximately \(1.4 \, mg\) of phosphorus-32 will remain.

Step by step solution

01

List the given information

We are given the following information: - Initial amount of phosphorus-32 (A₀) = 3.5 mg - Half-life (half_life) = 14.29 days - Time passed (time) = 1 month = 30 days
02

Set up the formula

The radioactive decay formula is given by: Amount remaining (A) = A₀ × (1/2)^(time / half_life) Let's use the given information to plug into the formula.
03

Plug in the given values

Substituting the given values into the radioactive decay formula, we get: A = 3.5 × (1/2)^(30/14.29)
04

Calculate (1/2)^(time / half_life)

(1/2)^(30/14.29) can be calculated as: (1/2)^(2.1) ≈ 0.40
05

Calculate the remaining amount of phosphorus-32

Now that we have the value of (1/2)^(30/14.29), we can find the remaining amount of phosphorus-32 as follows: A ≈ 3.5 × 0.40 = 1.4 mg
06

Final Answer

After 1 month (30 days), approximately 1.4 mg of phosphorus-32 will remain.

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