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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?

Short Answer

Expert verified
(a) Alpha decay is possible; energy released is positive. (b) Proton emission is not energetically possible. (c) Neutron emission is energetically possible. (d) \(\beta^-\) decay is not possible. (e) \(\beta^+\) decay is not possible.

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.

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Most popular questions from this chapter

Many radioactive decays occur within a sequence of decays for example, \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\). The half-life for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay is \(2.46 \times 10^{5}\) y, and the half-life for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay is \(7.54 \times 10^{4}\) y. Let 1 refer to \(^{234}_{92}U\), 2 to \(^{230}_{88}Th\), and 3 to \(^{226}_{84}Ra\); let \(\lambda\)1 be the decay constant for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay and \(\lambda\)2 be the decay constant for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay. The amount of \(^{230}_{88}Th\) present at any time depends on the rate at which it is produced by the decay of \(^{234}_{92}U\) and the rate by which it is depleted by its decay to \(^{226}_{84}Ra\). Therefore, \(d$$N_{2}\)(\(t\))/\(d$$t\) = \(\lambda$$_1$$N$$_1\)(\(t\)) - \(\lambda$$_2$$N$$_2\)(\(t\)). If we start with a sample that contains \(N_{10}\) nuclei of \(^{234}_{92}U\) and nothing else, then \(N{(t)}\) = \(N_{01}\)e\(^{-\lambda_{1}t}\). Thus \(dN_{2}(t)/dt\) = \(\lambda\)1\(N_{01}\)e\(^{-\lambda_{1}t}\)- \(\lambda2N_{2}(t)\). This differential equation for \(N_{2}(t)\) can be solved as follows. Assume a trial solution of the form \(N_{2}(t)\) = \(N_{10} [h_{1}e^{-\lambda_1t}\) + \(h_{2}e^{-\lambda_2t}\)] , where \(h_{1}\) and \(h_{2}\) are constants. (a) Since \(N_{2}(0)\) = 0, what must be the relationship between \(h_{1}\) and \(h_{2}\)? (b) Use the trial solution to calculate \(dN_{2}(t)/dt\), and substitute that into the differential equation for \(N_{2}(t)\). Collect the coefficients of \(e^{-\lambda_{1}t}\) and \(e^{-\lambda_{2}t}\). Since the equation must hold at all t, each of these coefficients must be zero. Use this requirement to solve for \(h_{1}\) and thereby complete the determination of \(N_{2}(t)\). (c) At time \(t = 0\), you have a pure sample containing 30.0 g of \(^{234}_{92}U\) and nothing else. What mass of \(^{230}_{88}Th\) is present at time \(t = 2.46 \times 10^{5}\) y, the half-life for the \(^{234}_{92}U\) decay?

Measurements on a certain isotope tell you that the decay rate decreases from 8318 decays/min to 3091 decays/min in 4.00 days. What is the half-life of this isotope?

Radioactive isotopes used in cancer therapy have a "shelf-life," like pharmaceuticals used in chemotherapy. Just after it has been manufactured in a nuclear reactor, the activity of a sample of \(^6$$^0\)Co is 5000 Ci. When its activity falls below 3500 Ci, it is considered too weak a source to use in treatment. You work in the radiology department of a large hospital. One of these \(^6$$^0\)Co sources in your inventory was manufactured on October 6, 2011. It is now April 6, 2014. Is the source still usable? The half-life of \(^6$$^0\)Co is 5.271 years.

How many protons and how many neutrons are there in a nucleus of the most common isotope of (a) silicon, \(^{28}_{14}Si\); (b) rubidium, \(^{85}_{37}Rb\); (c) thallium, \(^{205}_{81}Tl\)?

\(^2$$^3$$^8\)U decays spontaneously by \(\alpha\) emission to \(^2$$^3$$^4\)Th. Calculate (a) the total energy released by this process and (b) the recoil velocity of the \(^2$$^3$$^4\)Th nucleus. The atomic masses are 238.050788 u for \(^2$$^3$$^8\)U and 234.043601 u for \(^2$$^3$$^4\)Th.

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