Chapter 36: Problem 59
(II) The americium nucleus, \({ }_{95}^{241} \mathrm{Am},\) decays to a neptunium nucleus, \({ }^{237} \mathrm{~Np},\) by emitting an alpha particle of mass \(4.00260 \mathrm{u}\) and kinetic energy \(5.5 \mathrm{MeV}\). Estimate the mass of the neptunium nucleus, ignoring its recoil, given that the americium mass is \(241.05682 \mathrm{u} .\)
Short Answer
Step by step solution
Define Alpha Decay Process
Write Mass-Energy Balance Equation
Rearrange to Solve for Neptunium Mass
Convert Kinetic Energy to Atomic Mass Units
Substitute Known Values
Compute Neptunium Mass
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Alpha Decay
- During alpha decay, the original nucleus, known as the parent, transforms into a new nucleus called the daughter, which is of a different element.
- For instance, in the example above, americium (\(^{241}_{95}\mathrm{Am}\)) decays to neptunium (\(^{237}_{93}\mathrm{Np}\)).
Mass-Energy Equivalence
- This principle states that mass can be transformed into energy and vice versa, implying they are two forms of the same thing.
- In nuclear reactions like alpha decay, a small amount of mass from the nucleus is converted into energy, contributing to the kinetic energy of the emitted particles.
Kinetic Energy
- The kinetic energy of the alpha particle is critical as it indicates the energy transferred from the decay to the particle movement.
- In the provided example, the alpha particle is given kinetic energy of 5.5 MeV during the decay of americium to neptunium.