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Chapter 24: Problem 33

What is the specific activity (in \(\mathrm{Ci} / \mathrm{g}\) ) if \(1.65 \mathrm{mg}\) of an isotope emits \(1.56 \times 10^{6} \alpha\) particles per second?

Short Answer

Expert verified
0.02555 Ci/g.

Step by step solution

01

- Convert emissions to disintegrations per second

The first step is to recognize that each emission of an \( \alpha \) particle corresponds to a disintegration. Given that the isotope emits \(1.56 \times 10^{6} \) \( \alpha \) particles per second, this value already represents the disintegrations per second (dps).
02

- Convert dps to Curies

Next, convert the disintegrations per second (dps) to Curies (Ci). By definition, \ 1 \ \( \text{Ci} \) is \( 3.7 \times 10^{10} \) disintegrations per second. The activity in Ci can thus be calculated using the formula: \[ \text{Activity (Ci)} = \frac{1.56 \times 10^{6} \, \text{dps}}{3.7 \times 10^{10} \, \text{dps/Ci}} \]
03

- Calculate the specific activity

The specific activity is given by the activity per unit mass of the isotope. First, convert the mass from mg to grams: \[ 1.65 \, \text{mg} = 1.65 \times 10^{-3} \, \text{g} \] Then, calculate the specific activity using the activity in Ci from Step 2 and the mass in grams: \[ \text{Specific Activity} = \frac{\text{Activity (Ci)}}{1.65 \times 10^{-3} \, \text{g}} \]
04

- Perform the final calculations

Putting it all together: \[ \text{Activity (Ci)} = \frac{1.56 \times 10^{6}}{3.7 \times 10^{10}} = 4.216 \times 10^{-5} \, \text{Ci} \] And then calculate the specific activity: \[ \text{Specific Activity} = \frac{4.216 \times 10^{-5}}{1.65 \times 10^{-3}} = 0.02555 \, \text{Ci/g} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

disintegration per second
When studying radioactive materials, we talk about disintegrations per second (dps) to measure how many atomic nuclei are breaking apart each second. Every time an isotope emits a particle, like an alpha particle, it signifies a disintegration. So, if we know how many alpha particles are emitted per second (as in our problem, where it's 1.56 million), we know the number of disintegrations per second. That's our starting point for calculating activity.
Curies conversion
The Curie (Ci) is a unit used to measure radioactive activity. One Curie equals 3.7 x 10^10 disintegrations per second. This is a very large number, so usually, our source will not be in whole Curies, but in smaller fractions. To convert disintegrations per second to Curies, divide the dps by 3.7 x 10^10. In this problem, we have 1.56 million dps. Doing the math, that equals:

\[\frac{1.56 \times 10^6 \text{dps}}{3.7 \times 10^{10} \text{dps/Ci}} = 4.216 \times 10^{-5} \text{Ci}\]

This value is our activity in Curies.
mass conversion
To find specific activity, we need the mass of the radioactive substance. The specific activity is measured in Curie per gram (Ci/g). First, convert mass from milligrams (mg) to grams (g). In our problem, 1.65 mg is equal to 1.65 x 10^(-3) g.

Finally, we calculate the specific activity using the formula:
\[\text{Specific Activity} = \frac{4.216 \times 10^{-5} \text{Ci}}{1.65 \times 10^{-3} \text{g}} = 0.02555 \text{Ci/g}\]

Thus, for 1.65 mg of our isotope emitting 1.56 million alpha particles per second, the specific activity is 0.02555 Ci/g.

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Most popular questions from this chapter

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