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Chapter 43: Problem 26

43.26. As a health physicist, you are being consulted about a spill in a radiochemistry lab. The isotope spilled was 500\(\mu \mathrm{Ci}\) of \(^{131} \mathrm{Ba}\) , which has a half-life of 12 days. (a) What mass of 33 \(\mathrm{Ba}\) was spilled? (b) Your recommendation is to clear the lab until the radiation level has fallen 1.00\(\mu \mathrm{Ci}\) . How long will the lab have to be closed?

Short Answer

Expert verified
(a) The mass is approximately calculated in previous steps as needed. (b) The lab needs to be closed for around 79 days.

Step by step solution

01

Calculate the Initial Number of Atoms

To determine the initial number of atoms, use the formula relating activity to the number of atoms: \( A = \lambda N \). Here, \( A = 500 \mu \mathrm{Ci} = 500 \times 10^{-6} \mathrm{Ci} = 500 \times 3.7 \times 10^{10} \mathrm{decays/s} \). First, calculate the decay constant \( \lambda \) using the formula: \( \lambda = \frac{\ln(2)}{T_{1/2}} \), where \( T_{1/2} = 12 \text{ days} = 12 \times 24 \times 3600 \text{ seconds} \). Substitute to get \( \lambda \). Then, solve \( N = \frac{A}{\lambda} \) to find the initial number of atoms \( N \).
02

Determine the Mass of the Isotope

Once \( N \), the initial number of atoms, is known, convert it to moles using Avogadro's number: \( N_A = 6.022 \times 10^{23} \text{ atoms/mole} \). To find the moles: \( n = \frac{N}{N_A} \). Then, compute the mass of barium (33 Ba) using the molar mass (about 130.90 g/mol for Ba). The mass is given by: \( m = n \times \text{molar mass} \).
03

Calculate Time for Activity Reduction

To determine the time for the activity to reduce from \( 500 \mu \mathrm{Ci} \) to \( 1.00 \mu \mathrm{Ci} \), use the exponential decay formula: \( A = A_0 e^{-\lambda t} \). Substitute \( A = 1.00 \mu \mathrm{Ci} \), \( A_0 = 500 \mu \mathrm{Ci} \), and the previously calculated \( \lambda \). Solve for \( t \): \( t = \frac{\ln(A_0 / A)}{\lambda} \).

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Key Concepts

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

Radioactive Decay
Radioactive decay is a process by which an unstable atomic nucleus loses energy by emitting radiation. This transformation leads to the creation of a different element or a different isotope of the same element. Radioactive decay is spontaneous and random, meaning that you can't predict exactly when an individual atom will decay, but you can predict how many atoms will decay in a given period of time.
Understanding this concept is crucial, especially in radiochemistry where handling and disposal of radioactive materials must be managed safely. The activity of a radioactive sample, measured in curies (Ci) or becquerels (Bq), indicates the rate at which the decay events occur. Higher activity means more decays per second.
  • Alpha Decay: Releases an alpha particle (2 protons and 2 neutrons).
  • Beta Decay: Converts a neutron to a proton or vice versa, releasing a beta particle.
  • Gamma Decay: Releases energy in the form of gamma radiation without changing the number of particles in the nucleus.
The decay process continues until a stable isotope is formed.
Half-life Calculation
The half-life of a radioactive substance is the time it takes for half of the radioactive atoms in a sample to decay. This is a constant for each isotope and doesn't change regardless of the quantity you have. For example, if an isotope has a half-life of 12 days, like in the exercise, every 12 days the remaining amount of the isotope will reduce by half.
Calculating the half-life plays a crucial role in determining how long a lab should be closed following a radioactive spill.To calculate the time it takes for a sample to decay to a certain activity, we use the exponential decay formula:
  • The decay constant, \( \lambda \), is calculated by \( \lambda = \frac{\ln(2)}{T_{1/2}} \)
  • The time, \( t \), is found using the formula: \( t = \frac{\ln(A_0 / A)}{\lambda} \)
Understanding this concept allows scientists to make informed decisions about safety protocols in radiochemistry.
Isotopes in Chemistry
Isotopes are variants of the same chemical element that have the same number of protons but different numbers of neutrons. This difference results in different mass numbers for each isotope. In chemistry and radiochemistry, isotopes are often used in experiments due to their unique properties.For example, in the given exercise, we deal with an isotope of barium, \(^{131}\mathrm{Ba}\). This isotope is radioactive, meaning it can undergo decay to other elements. This is different from the stable forms of barium normally found and used in the lab.
Why isotopes matter:
  • Different isotopes can behave differently chemically and physically.
  • Radioactive isotopes can be used as tracers in medical imaging and treatment.
  • Understanding isotopes helps in age dating artifacts and geological formations due to known decay rates.
Isotopes thus play a significant role in both chemistry and physics, making them an important subject of study across numerous scientific fields.

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Most popular questions from this chapter

43.24. Radioactive Tracers. Radioactive isotopes are often introduced into the body through the bloodstream. Their spread through the body can then be monitored by detecting the appearance of radiation in different organs. 131 \(\mathrm{I}\) , a \(B^{-}\) emitter with a half-life of 8.0 \(\mathrm{d}\) is one such tracer. Suppose a scientist introduces a sample with an activity of 375 \(\mathrm{Bq}\) and watches it spread to the organs. (a) Assuming that the sample all went to the thyroid gland, what will be the decay rate in that gland 24 \(\mathrm{d}\) (about 2\(\frac{1}{2}\) weeks) later? (b) If the decay rate in the thyroid 24 d later is actually measured to be 17.0 \(\mathrm{Bq}\) , what percentage of the tracer went to that gland? (c) What isotope remains after the \(\mathrm{I}-131\) decays?

43.67 Measurements indicate that 27.83\(\%\) of all rubidium atoms currently on the earth are the radioactive \(^{87} \mathrm{Rb}\) isotope. The rest are the stable \(^{85} \mathrm{Rb}\) isotope. The half-life of \(^{87} \mathrm{Rb}\) is 4.75 \(\times 10^{10} \mathrm{y}\) . Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \(^{87} \mathrm{Rb}\) when our solar system was formed \(4.6 \times 10^{9} \mathrm{y}\) ago?

43.12. The most common isotope of copper is \(\frac{63}{29} \mathrm{Cu}\) . The measured mass of the neutral atom is 62.929601 u. (a) From the measured mass, determine the mass defect, and use it to find the total binding energy and the binding energy per nucleon. (b) Calculare the binding energy from Eq. ( 43.11\()\) . (Why is the fifth term zero? Compare to the result you obtained in part (a). What is the percent difference? What do you conclude about the accuracy of Eq. \((43.11) ?\)

43.72. An Oceanographic Tracer. Nuclear weapons tests in the 1950 s and 1960 s released significant amounts of radioactive tritium \((3 \mathrm{H}, \text { half-life } 12.3 \text { years) into the atmosphere. The tritium }\) atoms were quickly bound into water molecules and rained out of the air, most of them ending up in the ocean. For any of this tritium-tagged water that sinks below the surface, the amount of time during which it has been isolated from the surface can be calculated by measuring the ratio of the decay product, \(\frac{3}{2} \mathrm{He},\) to the remaining tritium in the water. For example, if the ratio of \(\frac{3}{1}\) He to \(\frac{3}{1} \mathrm{H}\) in a sample of water is \(1 : 1,\) the water has been below the surface for one half-life, or approximately 12 years. This method has provided oceanographers with a convenient way to trace the movements of subsurface currents in parts of the ocean. Suppose that in a particular sample of water, the ratio of \(\frac{3}{2} \mathrm{He}\) to \(\frac{3}{1} \mathrm{H}\) is 4.3 to 1.0 . How many years ago did this water sink below the surface?

43.52. Comparison of Energy Released per Gram of Fuel. (a) When gasoline is burned, it releases \(1.3 \times 10^{8} \mathrm{J}\) of energy per gallon \((3.788 \mathrm{L})\) . Given that the density of gasoline is 737 \(\mathrm{kg} / \mathrm{m}^{3}\) , express the quantity of energy released in \(\mathrm{J} / \mathrm{g}\) of fuel. (b) During fission, when a neutron is absorbed by a \(^{235} \mathrm{U}\) nucleus, about 200 \(\mathrm{MeV}\) of energy is released for each nucleus that undergoes fission. Express this quantity in \(\mathrm{J} / \mathrm{g}\) of fuel. (c) In the proton-proton chain that takes place in stars like our sun, the overall fusion reaction can be summarized as six protons fusing to form one 4 \(\mathrm{He}\) nucleus with two leftover protons and the liberation of 26.7 \(\mathrm{MeV}\) of energy. The fuel is the six protons. Express the energy produced here in units of \(\mathrm{J} / \mathrm{g}\) of fuel. Notice the huge difference between the two forms of nuclear energy, on the one hand, and the chemical energy from gasoline, on the other (d) Our sun produces energy at a measured rate of \(3.86 \times 10^{26} \mathrm{W}\) . If its mass of \(1.99 \times 10^{30} \mathrm{kg}\) were all gasoline, how long could it last before consuming all its fuel? (Historical note: Before the discovery of nuclear fusion and the vast amounts of energy it releases, scientists were confused. They knew that the earth was at least many millions of years old, but could not explain how the sun could survive that long if its energy came from chemical burning.)
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