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The polonium isotope \(^{210}_{84}Po\) has atomic mass 209.982874 u. Other atomic masses are \(^{206}_{82}Pb\), 205.974465 u; \(^{209}_{83}Bi\), 208.980399 u; \(^{210}_{83}Bi\), 209.984120 u; \(^{209}_{84}Po\), 208.982430 u; and \(^{210}_{85}At\), 209.987148 u. (a) Show that the alpha decay of \(^{210}_{84}Po\) is energetically possible, and find the energy of the emitted a particle. (b) Is \(^{210}_{84}Po\) energetically stable with respect to emission of a proton? Why or why not? (c) Is \(^{210}_{84}Po\) energetically stable with respect to emission of a neutron? Why or why not? (d) Is \(^{210}_{84}Po\) energetically stable with respect to \(\beta$$^-\) decay? Why or why not? (e) Is \(^{210}_{84}Po\)energetically stable with respect to \(\beta$$^+\) decay? Why or why not?

Step by step solution

01

Determine Alpha Decay Products

In alpha decay, a nucleus emits an alpha particle, which is a helium nucleus with 2 protons and 2 neutrons. For \(^{210}_{84}Po\), the resulting nucleus from alpha decay is \(^{206}_{82}Pb\) (lead), as calculated from: \(^{210}_{84}Po \rightarrow ^{206}_{82}Pb + ^{4}_{2}He\).

02

Calculate Mass Difference for Alpha Decay

Use the atomic masses to calculate the mass defect. The mass of \(^{210}_{84}Po\) is 209.982874 u, and for \(^{206}_{82}Pb\), it is 205.974465 u. The mass of the alpha particle \(^4_2He\) is approximately 4.00260 u. Calculate the mass defect as: \(209.982874 \text{ u} - (205.974465 \text{ u} + 4.00260 \text{ u})\).

03

Energy Calculation from Mass Defect

Convert the mass defect into energy using Einstein's equation \(E = \Delta m \cdot c^2\), where \(c\) is the speed of light. The conversion factor from atomic mass units to energy is 931.5 MeV/u. Calculate the energy released by \(\Delta E = \Delta m \times 931.5 \text{ MeV/u}\). Since the mass difference \(\Delta m\) is positive, alpha decay is energetically possible.

04

Check Stability for Proton Emission

Consider a hypothetical decay of \(^{210}_{84}Po\) into \(^{209}_{83}Bi\) by emission of a proton. Compare the mass of \(^{210}_{84}Po\) (209.982874 u) with the sum of the masses of \(^{209}_{83}Bi\) (208.980399 u) and a proton (approximately 1.007825 u). Calculate the difference. If the mass of \(^{210}_{84}Po\) is greater, the decay is energetically possible.

05

Check Stability for Neutron Emission

Consider the hypothetical neutron emission from \(^{210}_{84}Po\) to form \(^{209}_{84}Po\). Compare the mass of \(^{210}_{84}Po\) (209.982874 u) with the sum of the masses of \(^{209}_{84}Po\) (208.982430 u) and a neutron (approximately 1.008665 u). Calculate the difference.

06

Evaluate \(\beta^-\) Decay Possibility

For \(\beta^-\) decay, \(^{210}_{84}Po\) would transition to \(^{210}_{85}At\), accompanied by an electron. Compare the mass of \(^{210}_{84}Po\) (209.982874 u) with \(^{210}_{85}At\) (209.987148 u). Calculate the difference. If the resulting mass is less, \(\beta^-\) decay is possible.

07

Evaluate \(\beta^+\) Decay Possibility

Consider \(\beta^+\) decay where \(^{210}_{84}Po\) transitions to \(^{210}_{83}Bi\), along with a positron. Compare the mass of \(^{210}_{84}Po\) with \(^{210}_{83}Bi\) plus two electron masses (since a positron is an anti-electron). If the original mass is greater, \(\beta^+\) decay would be possible, otherwise not.

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Key Concepts

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

Nuclear Stability

When we talk about nuclear stability, we're considering how likely it is for a nucleus to stay together without disintegrating. For a nucleus to be stable, the forces holding it together must balance the forces trying to pull it apart. In isotopes like polonium-210, these forces are the strong nuclear force and the electromagnetic force.

Factors that affect nuclear stability include:

• Proton-to-neutron ratio: A balanced ratio helps ensure stability.
• Binding energy: Higher binding energy per nucleon usually means a more stable nucleus.
• Decay potential: Nuclei with too many protons or neutrons may emit particles to reach a stable state.

Understanding these aspects is crucial for determining if processes like alpha decay, beta decay, or neutron/proton emissions are possible.

Mass Defect

Mass defect is a fascinating concept that arises because the mass of an atomic nucleus is actually less than the sum of the individual masses of its protons and neutrons. This discrepancy is what we call the mass defect.

Here's how it works:

• During the formation of a nucleus, some mass is converted into energy, which binds the protons and neutrons together.
• This missing mass, or mass defect, accounts for the energy required to separate a nucleus into its individual nucleons, known as binding energy.
• When calculating whether nuclear decay is possible, we need to evaluate the mass defect to understand the energy releases that might occur.

In alpha decay, assessing the mass defect helps us predict energy changes during the decay process.

Beta Decay

Beta decay is a transformation within the nucleus, where a neutron converts into a proton (or vice versa), resulting in the emission of a beta particle. There are two main types:

• Beta-minus decay: Involves the conversion of a neutron to a proton, accompanied by the emission of an electron and an antineutrino. This type of decay increases the atomic number by one.
• Beta-plus decay: Occurs when a proton transforms into a neutron, emitting a positron and a neutrino, which decreases the atomic number by one.

In the context of polonium-210, evaluating its ability to undergo beta decay involves comparing the initial and potential decay products' masses. This assessment determines if the decay is energetically possible and whether the new nucleus is more stable than the original.

Energy Calculation

Energy calculations in nuclear reactions are critical to understanding whether a particular decay process will occur. In nuclear physics, the mantra "mass equals energy" is often realized through Einstein's equation: \[E = \Delta m \cdot c^2\]Where \(E\) is the energy released, \(\Delta m\) is the mass defect, and \(c\) is the speed of light. In practical applications, we convert mass change into energy using a constant, 931.5 MeV/u.

To determine if a decay process like alpha decay is energetically feasible, we:

• Calculate the total mass of both the parent and potential decay products.
• Find the mass defect by subtracting the total mass of products from the total mass of the original nucleus.
• Convert this mass defect into energy to see if it is positive, which implies the release of energy, indicating the decay is possible.

This approach helps scientists and students understand the energetic pathways of nuclear transformations.