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Ch 43: Nuclear Physics
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 43: Nuclear PhysicsProblem 37
Chapter 43, Problem 37

A 6767-kg person accidentally ingests 0.350.35 Ci of tritium.
(a) Assume that the tritium spreads uniformly throughout the body and that each decay leads on the average to the absorption of 5.05.0 keV of energy from the electrons emitted in the decay. The half-life of tritium is 12.312.3 y, and the RBE of the electrons is 1.01.0. Calculate the absorbed dose in rad and the equivalent dose in rem during one week.
(b) The β−\(\beta\)^{-} decay of tritium releases more than 5.05.0 keV of energy. Why is the average energy absorbed less than the total energy released in the decay?

Verified step by step guidance
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Step 1: Understand the problem and identify the given data. The person has ingested 0.35 Ci of tritium, which spreads uniformly throughout their body. Each decay releases 5.0 keV of energy absorbed by the body. The half-life of tritium is 12.3 years, and the RBE (Relative Biological Effectiveness) is 1.0. We need to calculate the absorbed dose in rad and the equivalent dose in rem for one week. Additionally, we need to explain why the average energy absorbed is less than the total energy released in the decay.
Step 2: Convert the activity from curies (Ci) to disintegrations per second (Bq). Use the conversion factor: 1 Ci = 3.7 × 10^10 Bq. Multiply the given activity (0.35 Ci) by this factor to find the activity in Bq.
Step 3: Calculate the total number of decays in one week. Use the formula for activity: \( N = A \cdot t \), where \( A \) is the activity in Bq and \( t \) is the time in seconds. Convert one week into seconds (1 week = 7 days × 24 hours/day × 3600 seconds/hour). Multiply the activity by the time to find the total number of decays.
Step 4: Calculate the total energy absorbed by the body. Multiply the total number of decays by the average energy absorbed per decay (5.0 keV). Convert the energy from keV to joules using the conversion factor: 1 eV = 1.602 × 10^-19 J. Then, divide the total energy by the mass of the person (67 kg) to find the absorbed dose in rad. Note: 1 rad = 0.01 J/kg.
Step 5: Calculate the equivalent dose in rem. Multiply the absorbed dose in rad by the RBE (1.0 in this case). Finally, for part (b), explain that the average energy absorbed is less than the total energy released because not all the energy from the decay is absorbed by the body. Some energy is carried away by neutrinos or escapes the body as radiation.

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Key Concepts

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

Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This can occur in various forms, including alpha, beta, and gamma decay. In the case of tritium, beta decay occurs, where a neutron is converted into a proton, emitting an electron and an antineutrino. Understanding this process is crucial for calculating the energy released and the biological effects of radiation exposure.
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Absorbed Dose and Equivalent Dose

The absorbed dose is a measure of the energy deposited by ionizing radiation in a given mass of tissue, typically expressed in grays (Gy) or rads. The equivalent dose accounts for the biological effect of the radiation type, using a quality factor known as the Relative Biological Effectiveness (RBE). This is expressed in sieverts (Sv) or rems. These concepts are essential for assessing the potential health risks associated with radiation exposure.
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Energy Transfer in Radiation

When radiation interacts with matter, not all the energy released during decay is absorbed by the surrounding tissue. Factors such as the range of the emitted particles, their energy, and the material properties influence how much energy is actually transferred. In the case of tritium's beta decay, while the total energy released is higher, the average energy absorbed is lower due to these interactions, which must be considered in dose calculations.
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Related Practice
Textbook Question

In a diagnostic x-ray procedure, 5.00×10105.00\(\times\)10^{10} photons are absorbed by tissue with a mass of 0.6000.600 kg. The x-ray wavelength is 0.0200 0.0200 nm.

(a) What is the total energy absorbed by the tissue?

(b) What is the equivalent dose in rem?

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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

Calculate the energy released in the fusion reaction: 23He+12H→24He+11H_2^3He+_1^2H\(\rightarrow\)_2^4He+_1^1H

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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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