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Chapter 21: Problem 39

Radium-226, which undergoes alpha decay, has a half-life of 1600 yr. (a) How many alpha particles are emitted in 5.0 min by a 10.0 -mg sample of \(^{226} \mathrm{Ra}\) ? (b) What is the activity of the sample in mCi?

Short Answer

Expert verified
(a) The number of alpha particles emitted in 5 minutes by a 10 mg sample of Radium-226 is found by calculating the number of decayed nuclei (\(N_{decay}\)) using the decay formula: \[ N_{decay} = N \times (1 - e^{-λ \times (9.512 \times 10^{-6}\, \text{years})}) \] (b) The activity of the sample in millicuries (mCi) can be calculated by converting the activity in Becquerels (Bq) to millicuries using the conversion factor: \[ \text{Activity} (\text{mCi}) = \frac{\text{Activity} (\text{Bq})}{3.7 \times 10^{10}\, \frac{\text{Bq}}{\text{Ci}}} × 10^{3}\, \frac{\text{mCi}}{\text{Ci}} \]

Step by step solution

01

Find the Decay Constant

Using the half-life formula, we can find the decay constant (λ) for Radium-226. Half-life formula: \[ T_{1/2} = \frac{0.693}{λ} \] Given half-life of Radium-226: \[ T_{1/2} = 1600\, \text{years} \] Calculate the decay constant: \[ λ = \frac{0.693}{T_{1/2}} \]
02

Convert Time to the Same Unit

To use the decay constant in our calculations, we need to convert the given time of 5 minutes into the same unit as the half-life (years). \[ 5.0\, \text{min} × \frac{1\, \text{hour}}{60\, \text{min}} × \frac{1\, \text{day}}{24\, \text{hours}} × \frac{1\, \text{year}}{365.25\, \text{days}} = 9.512 \times 10^{-6}\, \text{years} \]
03

Find the Number of Nuclei in the Sample

Now we need to find the number of Radium-226 nuclei in the 10 mg sample. Given sample mass: \[ m = 10\, \text{mg} \] Mass of one Radium-226 nucleus: \[ m_{Ra} = \frac{226\, \text{g/mol}}{(6.022 \times 10^{23}\, \text{nuclei/mol})} \] Now, calculate the number of Radium-226 nuclei (\(N\)) in the sample: \[ N = \frac{m}{m_{Ra}} \]
04

Find the Number of Nuclei That Decay in 5 Minutes

Using the decay formula, we can find the number of nuclei that decay in 5 minutes: Decay Formula: \[ N_{decay} = N \times (1 - e^{-λ \times t}) \] Plug in the values from previous steps: \[ N_{decay} = N \times(1 - e^{-λ \times (9.512 \times 10^{-6}\, \text{years})}) \]
05

Convert the Number of Decayed Nuclei to Alpha Particles

Since one alpha particle is emitted per decay, the number of alpha particles is equal to the number of decayed nuclei: \[ \text{Alpha particles} = N_{decay} \]
06

Calculate the Activity in Becquerels

Now, we can find the activity of the sample by dividing the number of alpha particles by the elapsed time of 5 minutes. First, we convert 5 minutes to seconds: \[ 5.0\, \text{min} × 60\, \frac{\text{s}}{\text{min}} = 300\, \text{s} \] Next, calculate the activity in Becquerels (Bq): \[ \text{Activity} = \frac{\text{Alpha particles}}{300\, \text{s}} \]
07

Convert the Activity to millicuries (mCi)

Finally, we convert the activity from Becquerels to millicuries: Conversion factor: \[ 1\, \text{Curie (Ci)} = 3.7 \times 10^{10}\, \text{Bq} \] Convert to millicuries: \[ \text{Activity} (\text{mCi}) = \frac{\text{Activity} (\text{Bq})}{3.7 \times 10^{10}\, \frac{\text{Bq}}{\text{Ci}}} × 10^{3}\, \frac{\text{mCi}}{\text{Ci}} \] By following the above steps, you will find the number of alpha particles emitted in 5 minutes by the 10 mg sample of Radium-226 and its activity in millicuries.

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Most popular questions from this chapter

Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of 8.02 days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of Nal, in which only a small fraction of the iodide is radioactive. (a) Why is Nal a good choice for the source of iodine? (b) If a Geiger counter is placed near the person's thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thyroid will take up about 12\(\%\) of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to 0.01\(\%\) of the original amount?

According to current regulations, the maximum permissible dose of strontium-90 in the body of an adult is 1\(\mu \mathrm{Ci}\left(1 \times 10^{-6} \mathrm{Ci}\right) .\) Using the relationship rate \(=k N,\) calculate the number of atoms of strontium-90 to which this dose corresponds. To what mass of strontium-90 does this correspond? The half-life for strontium-90 is 28.8 yr.

Chlorine has two stable nuclides, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl} .\) In contrast, \(^{36} \mathrm{Cl}\) is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of \(^{36} \mathrm{Cl} ?\) (b) Based on the empirical rules about nuclear stability, explain why the nucleus of \(^{36} \mathrm{C}\) is less stable than either \(^{35}\mathrm{Cl}\) or \(^{37} \mathrm{Cl}\).

Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

A wooden artifact from a Chinese temple has a \(^{14} \mathrm{C}\) cactivity of 38.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. From the half-life for \(^{14} \mathrm{C}\) decay, 5715 yr, determine the age of the artifact.
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