/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 As a health physicist, you are b... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

As a health physicist, you are being consulted about a spill in a radiochemistry lab. The isotope spilled was \(400 \mu \mathrm{Ci}\) of \({ }^{131} \mathrm{Ba},\) which has a half-life of 12 days. (a) What mass of \({ }^{131}\) 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
The mass of Ba-131 that was spilled is approximately 1.25 micrograms. The lab needs to be closed for approximately 37 days.

Step by step solution

01

Calculation of the amount of Ba-131

First, calculate the number of atoms using the given activity. The activity A of a radioactive substance is given by \(A = \lambda * N\), where \(N\) is the number of atoms of the substance and \(\lambda = \frac{ln(2)}{t_{1/2}}\) is the decay constant. Here, \( t_{1/2} \) is the half-life of the substance. So, the number of atoms \(N\) can be found out by manipulating the above equation as \(N = \frac{A}{\lambda}\). Substituting the given values, we can calculate \(N\). \(N\) then can be used to find out the mass \(m\) in grams of Ba-131 by using the equation \(N = \frac{m}{M}\) where \(M\) is the Molar mass of Ba-131 (approximately 131 g/mol). Solving this equation gives us the mass \(m\).
02

Calculation of the required time for safe levels

The half-life formula is used to calculate the time required for the radiation level to fall to 1 micro Ci. The formula is \(A = A_{0} * e^{-\lambda t}\) where \(A_{0}\) is the initial level (400 micro Ci), \(\lambda\) is the decay constant, \(A\) is the final level (1 micro Ci) and \(t\) is the time required for this reduction. Solving this equation for \(t\) gives us the required time. Convert this time to days for the final answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Decay Constant
In the world of radioactive substances, understanding the decay constant is pivotal. The decay constant, often represented by the Greek letter lambda (\(\lambda\)), signifies the probability per unit time that a particular atom will decay. It helps to describe how fast a radioactive isotope will decay over time.
To determine \(\lambda\), we use the relation \(\lambda = \frac{\ln(2)}{t_{1/2}}\), where \(t_{1/2}\) is the half-life of the substance. This equation links the decay constant directly to half-life, providing insights into the decay process. The decay constant is essential for calculating the number of remaining atoms at a future time, using the equation \(A = \lambda N\), where \(A\) is the activity of the source, and \(N\) is the number of undecayed atoms. It plays an important role in understanding how long hazardous materials will remain active.
Half-Life
Half-life is a crucial concept when discussing radioactive decay. It refers to the time required for half of the radioactive atoms in a sample to decay. In simpler terms, it tells us how quickly a substance loses its radioactive potency. For \(^{131}\mathrm{Ba}\), the half-life is 12 days.
During each half-life, the quantity of the radioactive substance falls by half. Knowing the half-life helps predict when a material will reach a safe level of radioactivity. For example, if an isotope starts at a particular activity, the time taken to reduce to a quarter of its initial activity will be two half-lives. This property allows scientists to estimate how long it will take for dangerous radiation to reduce to safer levels, making half-life an essential factor in radiation safety assessments.
Radiation Safety
Ensuring radiation safety is paramount when dealing with radioactive substances. When a spill occurs, as in a radiochemistry lab, it is crucial to calculate how long it will take for the radiation levels to fall to a safe threshold.
Using the equation \(A = A_0 e^{-\lambda t}\), where \(A_0\) is the initial activity, \(A\) is the desired final activity, \(\lambda\) is the decay constant, and \(t\) is the time elapsed, we can determine when the environment will be safe. It's important to close off the area until the radiation decays to the recommended safety level (often a significant drop from the original level). By calculating the proper duration for exclusion, health physicists can ensure the wellbeing of individuals by minimizing exposure to hazardous radiation.
Molar Mass
Molar mass, denoted as \(M\), is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For \(^{131}\mathrm{Ba}\), the molar mass is approximately 131 g/mol, which is important when converting between the number of atoms and mass.
To find the mass of a radioactive substance using its molar mass, we use the equation \(N = \frac{m}{M}\), where \(N\) is the number of atoms and \(m\) is the mass in grams. In the context of radioactive spills, calculating the molar mass helps in determining how much of a substance has been released. Molar mass also aids in converting from the microcuries of radioactivity to the actual physical quantity of the spilled material, aligning the scientific calculations with practical environmental safety considerations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Thorium \({ }_{90}^{230} \mathrm{Th}\) decays to radium \({ }_{88}^{226} \mathrm{Ra}\) by \(\alpha\) emission. The masses of the neutral atoms are 230.033134 u for \({ }_{90}^{230} \mathrm{Th}\) and \(226.025410 \mathrm{u}\) for \({ }_{88}^{226} \mathrm{Ra}\). If the parent thorium nucleus is at rest, what is the kinetic energy of the emitted \(\alpha\) particle? (Be sure to account for the recoil of the daughter nucleus.)

Consider the nuclear reaction $$ { }_{2}^{4} \mathrm{He}+{ }_{3}^{7} \mathrm{Li} \rightarrow \mathrm{X}+{ }_{0}^{1} \mathrm{n} $$ where \(X\) is a nuclide. (a) What are \(Z\) and \(A\) for the nuclide \(X ?\) (b) Is energy absorbed or liberated? How much?

The sun produces a minority \(16.68 \%\) of its helium-4 using the p-p II chain, in which four protons and an ambient alpha particle produce two alpha particles, two neutrinos, and at least four gamma-ray photons. The first two steps are the same as the first two steps of the p-p I chain (see Problem 43.62). However, the third step is \({ }_{2}^{3} \mathrm{He}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{4}^{7} \mathrm{Be}+\gamma,\) where \({ }_{4}^{7} \mathrm{Be}\) is an unstable beryllium nuclide with mass 7.016930 u. (a) What is the energy of the photon? (b) The beryllium nuclide then captures an electron, which changes one of its protons into a neutron, resulting in \({ }_{4}^{7} \mathrm{Be}+e^{-} \rightarrow{ }_{3}^{7} \mathrm{Li}+v_{e},\) where \({ }_{3}^{7} \mathrm{Li}\) is the common (and stable) nuclide of lithium, with mass \(7.016003 \mathrm{u}\). The excess energy is carried away by the neutrino. What is the energy of that neutrino? (c) A proton then collides with the lithium nucleus, resulting in two alpha particles via \({ }_{1}^{1} \mathrm{H}+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He} .\) How much energy is released in this step? (d) If each alpha particle has the same speed, then in the rest frame of the lithium disintegration, what is the speed of each alpha particle? (e) How much energy is released in total by the processes in this \(\mathrm{p}\) -p II chain? (f) How does this compare with the energy released in the p-p I chain? (g) What percentage of the energy in the \(\mathrm{p}\) -p II chain is carried away by neutrinos?

The carbon isotope \({ }_{6}^{11} \mathrm{C}\) undergoes \(\beta^{+}\) (positron) decay. The atomic mass of \({ }_{6}^{11} \mathrm{C}\) is \(11.011433 \mathrm{u}\). (a) How many protons and how many neutrons are in the daughter nucleus produced by this decay? (b) How much energy, in \(\mathrm{MeV},\) is released in the decay of one \({ }_{6}^{11} \mathrm{C}\) nucleus?

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. \((\mathrm{c})\) 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 \({ }^{4}\) 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.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.