Question

A scintillation counter registers emitted radiation caused by the disintegration of nuclides. If each atom of a nuclide emits one count, what is the activity of a sample that registers \(3.00 \times 10^{4}\) disintegrations in five minutes?

Short answer

Question: A sample of nuclide registers 3.00 x 10^4 disintegrations in five minutes. Calculate its activity in becquerels (Bq). Answer: The activity of the given sample is 1.00 x 10^2 becquerels (Bq).

Step-by-step solution

Convert time to seconds

First, convert the given time of five minutes into seconds. There are 60 seconds in a minute, so to convert five minutes into seconds, multiply 5 by 60. Time = 5 minutes × 60 seconds/minute

Calculate the activity in becquerels

The activity is the number of decays per unit time, so divide the number of disintegrations (\(3.00 \times 10^{4}\)) by the time in seconds, which we calculated in Step 1. Activity (in becquerels) = (No. of disintegrations) / (time in seconds)

Substitute given values and calculate the activity

Now, we will take the given number of disintegrations, \(3.00 \times 10^{4}\), and the time we calculated in the first step (5 minutes × 60 seconds/minute) and substitute them into the expression for activity, solving it. Activity (in becquerels) = \((3.00 \times 10^{4}) \ / \ (5 \times 60)\)

Simplify the expression and find the activity

Simplify the expression by performing the calculations: Activity (in becquerels) = \((3.00 \times 10^{4}) \ / \ (300)\) Activity (in becquerels) = \(1.00 \times 10^{2} \ \text{Bq}\) Hence, the activity of the given sample is \(1.00 \times 10^{2}\) becquerels.