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Chapter 43: Problem 70

A \(^{186}_{76} \mathrm{Os}\) nucleus at rest decays by the emission of a 2.76 -MeV \(\alpha\) particle. Calculate the atomic mass of the daughter nuclide produced by this decay, assuming that it is produced in its ground state. The atomic mass of \(^{186}_{76} \mathrm{Os}\) is 185.953838 \(\mathrm{u}\)

Short Answer

Expert verified
The atomic mass of the daughter nuclide \( ^{182}_{74} \mathrm{W} \) is 181.948272 u.

Step by step solution

01

Understand the Decay Process

In this exercise, an \( ^{186}_{76} \mathrm{Os} \) nucleus emits an \( \alpha \) particle, which consists of 2 protons and 2 neutrons, reducing the original nucleus by 4 nucleons total. This transforms \( ^{186}_{76} \mathrm{Os} \) (Osmium) into another element and changes its atomic and mass numbers.
02

Identify the Daughter Nucleus

When an \( ^{186}_{76} \mathrm{Os} \) nucleus emits an \( \alpha \) particle, the atomic number decreases by 2 (from 76 to 74) and the mass number decreases by 4 (from 186 to 182). Thus, the daughter nucleus is \( ^{182}_{74} \mathrm{W} \) (Tungsten).
03

Calculate the Mass of the Daughter Nuclide

To calculate the atomic mass of the daughter nuclide, we use the equation: \( m_d = m_0 - m_{\alpha} - \Delta E \), where \( m_d \) is the mass of the daughter nuclide, \( m_0 \) is the mass of the original nucleus, \( m_{\alpha} \) is the mass of the \( \alpha \) particle (approximately 4.002602 \( \mathrm{u} \)), and \( \Delta E \) is the energy released in the form of the \( \alpha \)-particle (2.76 MeV), converted into atomic mass units: \( \frac{2.76 \ \mathrm{MeV}}{931.5 \ \mathrm{MeV/u}} = 0.002964 \mathrm{u} \).
04

Perform the Calculation

Substitute the known values into the equation: \[ m_d = 185.953838 \ \mathrm{u} - 4.002602 \ \mathrm{u} - 0.002964 \ \mathrm{u} \] Simplify to find the atomic mass of the daughter nuclide: \[ m_d = 181.948272 \ \mathrm{u} \].

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Key Concepts

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

Atomic Mass
Atomic mass is a critical concept in understanding nuclear reactions like alpha decay. It represents the mass of an atomic nucleus and is expressed in atomic mass units (u), where one atomic mass unit is defined as one twelfth of the mass of a carbon-12 atom.

When an element undergoes alpha decay, it releases an alpha particle, which affects the atomic mass of the original element. The alpha particle consists of two protons and two neutrons, significantly changing the mass of the parent nucleus.

In the problem provided, the initial atomic mass of osmium (Os) is given as 185.953838 u. By subtracting the mass of the alpha particle and the energy released during the decay (expressed as mass), we find the atomic mass of the daughter nuclide.
  • Atomic mass is crucial for predicting changes in nuclear reactions.
  • It's measured in atomic mass units (u).
  • The mass of particles involved in nuclear reactions, like the alpha particle, alters the atomic mass of the resultant element.
Daughter Nuclide
A daughter nuclide is the new nucleus formed when a parent nucleus undergoes radioactive decay. During alpha decay, the parent nucleus emits an alpha particle and transforms into the daughter nuclide.

This process involves a shift in both atomic and mass numbers. An alpha particle, being composed of 2 protons and 2 neutrons, reduces the atomic number by 2 and the mass number by 4. In the provided exercise, when an osmium-186 nucleus emits an alpha particle, it transforms into a tungsten-182 nucleus.

Understanding the concept of daughter nuclides helps in tracking the changes in the elements through various nuclear reactions.
  • The daughter nuclide is a direct result of the decay process.
  • Alpha decay reduces atomic and mass numbers.
  • Daughter nuclides are major players in understanding the nature of radioactive decay.
Nuclear Reaction
Nuclear reactions involve changes in an atom's nucleus and are fundamental to concepts like alpha decay. During these reactions, nuclei collide or decay to form new elements.

Alpha decay specifically is a type of radioactive decay where an unstable nucleus releases an alpha particle, reducing its mass and transforming into a different element. This releasing involves the emission of 2 protons and 2 neutrons. These reactions significantly impact the stability of an element and the composition of the atomic nucleus.

  • Nuclear reactions change the nucleus of atoms.
  • Alpha decay is an example of a nuclear reaction.
  • Nuclear reactions can alter element identities and properties.
Mass-Energy Equivalence
The principle of mass-energy equivalence, introduced by Albert Einstein in his famous equation \( E=mc^2 \), is crucial in understanding nuclear reactions and decay processes.

This principle states that mass and energy are interchangeable. In nuclear reactions like alpha decay, the energy released can be calculated as a change in mass. When an alpha particle is emitted, not only is mass lost in the form of the particle, but energy is also released.

In the exercise, the energy output is calculated and converted into an equivalent mass loss to determine the atomic mass of the daughter nuclide. This underscores the interconnectedness of mass and energy in nuclear physics.
  • Mass-energy equivalence connects the dots between mass and energy.
  • Energy release during decay corresponds to a loss in mass.
  • This principle is essential in understanding and predicting nuclear behavior.

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Most popular questions from this chapter

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

We Are Stardust. In 1952 spectral lines of the element technetium- 99\(\left(^{99} \mathrm{Tc}\right)\) were discovered in a red giant star. Red giants are very old stars, often around 10 billion years old, and near the end of their lives. Technetium has \(no\) stable isotopes, and the half-life of \(^{99} \mathrm{Tc}\) is \(200,000\) years. (a) For how many half-lives has the \(^{99} \mathrm{Tc}\) been in the red-giant star if its age is 10 billion years? (b) What fraction of the original \(^{99} \mathrm{Tc}\) would be left at the end of that time? This discovery was extremely important because it provided convincing evidence for the theory (now essentially known to be true) that most of the atoms heavier than hydrogen and helium were made inside of stars by thermonuclear fusion and other nuclear processes. If the \(^{99} \mathrm{Tc}\) had been part of the star since it was born, the amount remaining after 10 billion years would have been so minute that it would not have been detectable. This knowledge is what led the late astronomer Carl Sagan to proclaim that "we are stardust."

BIO A 70.0-kg person experiences a whole-body exposure to \(\alpha\) radiation with energy 4.77 MeV. A total of \(6.25 \times 10^{12} \alpha\) particles are absorbed. (a) What is the absorbed dose in rad? (b) What is the equivalent dose in rem? (c) If the source is 0.0320 \(\mathrm{g}\) of \(^{226} \mathrm{Ra}\) (half-life 1600 \(\mathrm{y}\)) somewhere in the body, what is the activity of this source? (d) If all the alpha particles produced are absorbed, what time is required for this dose to be delivered?

BIO A person exposed to fast neutrons receives a radiation dose of 200 rem on part of his hand, affecting 25 g of tissue. The RBE of these neutrons is 10. (a) How many rad did he receive? (b) How many joules of energy did this person receive? (c) Suppose the person received the same rad dosage, but from beta rays with an RBE of 1.0 instead of neutrons. How many rem would he have received?

Tritium (\(^{3}_{1}\)H) is an unstable isotope of hydrogen; its mass, including one electron, is 3.016049 u. (a) Show that tritium must be unstable with respect to beta decay because the decay products ( \(^{3}_{2}\)He plus an emitted electron ) have less total mass than the tritium. \right. (b) Determine the total kinetic energy (in MeV) of the decay products, taking care to account for the electron masses correctly.
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