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Ch 43: Nuclear Physics
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 42, Problem 21

The common isotope of uranium, 238U^{238}U, has a half-life of 4.47×1094.47\(\times\)10^9 years, decaying to 234Th^{234}Th by alpha emission.
(a) What is the decay constant?
(b) What mass of uranium is required for an activity of 1.001.00 curie?
(c) How many alpha particles are emitted per second by 10.010.0 g of uranium?

Verified step by step guidance
1
Step 1: To find the decay constant (λ) in part (a), use the relationship between the half-life (T₁/₂) and the decay constant: λ = ln(2) / T₁/₂. Substitute the given half-life of uranium-238, T₁/₂ = 4.47 × 10⁹ years, into the formula. Ensure the units of time are consistent (e.g., convert years to seconds if needed).
Step 2: For part (b), use the relationship between activity (A), decay constant (λ), and the number of nuclei (N): A = λN. Rearrange to find N = A / λ. Then, use the relationship between the number of nuclei and the mass of uranium: N = (m / M) × Nₐ, where m is the mass of uranium, M is the molar mass of uranium-238 (238 g/mol), and Nₐ is Avogadro's number (6.022 × 10²³ nuclei/mol). Combine these equations to solve for the required mass m.
Step 3: For part (c), calculate the number of nuclei (N) in 10.0 g of uranium using the formula N = (m / M) × Nₐ, where m = 10.0 g, M = 238 g/mol, and Nₐ = 6.022 × 10²³ nuclei/mol. This gives the total number of uranium nuclei in the sample.
Step 4: Use the activity formula A = λN to calculate the activity of the 10.0 g uranium sample. Substitute the decay constant (λ) from part (a) and the number of nuclei (N) from step 3 into the formula to find the activity in disintegrations per second (Becquerels).
Step 5: Since each decay corresponds to the emission of one alpha particle, the number of alpha particles emitted per second is equal to the activity calculated in step 4. Express the result in terms of alpha particles per second.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Decay Constant

The decay constant (λ) is a probability rate at which a radioactive isotope decays. It is related to the half-life (t½) of the isotope by the formula λ = ln(2) / t½. For uranium-238, knowing the half-life allows us to calculate the decay constant, which is essential for determining the rate of decay and the activity of the sample.
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Radioactive Activity

Radioactive activity is the rate at which a sample of radioactive material decays, measured in disintegrations per second. The unit of activity is the curie (Ci), where 1 Ci equals 3.7 x 10^10 disintegrations per second. Understanding how to relate mass to activity is crucial for calculating how much uranium is needed to achieve a specific activity level.
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Alpha Emission

Alpha emission is a type of radioactive decay in which an atomic nucleus emits an alpha particle, consisting of two protons and two neutrons. This process reduces the mass number of the original nucleus by four and the atomic number by two, transforming the element into a different isotope. Knowing the nature of alpha decay is important for calculating the number of alpha particles emitted from a given mass of uranium.
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Related Practice
Textbook Question

What nuclide is produced in the following radioactive decays?

(a) α\(\alpha\) decay of 94239Pu_{94}^{239}Pu

(b) β\(\beta\)^{-} decay of 1124Na_{11}^{24}Na

(c) β+\(\beta\)^{+} decay of 815O_8^{15}O

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Textbook Question

Measurements on a certain isotope tell you that the decay rate decreases from 83188318 decays/min to 30913091 decays/min in 4.004.00 days. What is the half-life of this isotope?

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Textbook Question

What particle (a particle, electron, or positron) is emitted in the following radioactive decays?

(a) 1427Si1327Al_{14}^{27}Si\(\rightarrow\)_{13}^{27}Al

(b) 92238U90234Th_{92}^{238}U\(\rightarrow\)_{90}^{234}Th

(c) 3374As3474Se_{33}^{74}As\(\rightarrow\)_{34}^{74}Se

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Textbook Question

Radioactive isotopes used in cancer therapy have a 'shelf-life,' like pharmaceuticals used in chemotherapy. Just after it has been manufactured in a nuclear reactor, the activity of a sample of 60Co^{60}Co is 50005000 Ci. When its activity falls below 35003500 Ci, it is considered too weak a source to use in treatment. You work in the radiology department of a large hospital. One of these 60Co^{60}Co sources in your inventory was manufactured on October 6, 2011. It is now April 6, 2014. Is the source still usable? The half-life of 60Co^{60}Co is 5.2715.271 years.

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Textbook Question

It has become popular for some people to have yearly whole-body scans (CT scans, formerly called CAT scans) using x rays, just to see if they detect anything suspicious. A number of medical people have recently questioned the advisability of such scans, due in part to the radiation they impart. Typically, one such scan gives a dose of 1212 mSv, applied to the whole body. By contrast, a chest x ray typically administers 0.200.20 mSv to only 5.05.0 kg of tissue. How many chest x rays would deliver the same total amount of energy to the body of a 7575-kg person as one whole-body scan?

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Textbook Question

At an archeological site, a sample from timbers containing 500500 g of carbon provides 26902690 decays/min. What is the age of the sample?

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