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Chapter 43: Problem 13

What nuclide is produced in the following radioactive decays? (a) \(\alpha\) decay of \(^{239}_{94}Pu\); (b) \(\beta$$^-\) decay of \(^{24}_{11}Na\); (c) \(\beta$$^+\) decay of \(^{15}_{8}O\).

Short Answer

Expert verified
(a) Uranium-235, (b) Magnesium-24, (c) Nitrogen-15.

Step by step solution

01

Understanding Alpha Decay

In alpha decay, an alpha particle, which is essentially a helium nucleus with 2 protons and 2 neutrons, is emitted from the parent nucleus. This reduces the mass number by 4 and the atomic number by 2.
02

Alpha Decay of Plutonium-239

Considering the alpha decay of \(^{239}_{94}Pu\), the emitted alpha particle is \(^{4}_{2}He\). Therefore, the resulting nuclide is calculated as follows:\[\begin{align*}A_f &= A_i - 4 = 239 - 4 = 235\Z_f &= Z_i - 2 = 94 - 2 = 92\end{align*}\]The product is \(^{235}_{92}U\) (Uranium-235).
03

Understanding Beta Minus Decay

In beta minus decay, a neutron is converted into a proton, and a beta particle (electron) is emitted. This increases the atomic number by 1, while the mass number remains unchanged.
04

Beta Minus Decay of Sodium-24

During the \(\beta^-\) decay of \(^{24}_{11}Na\), the atomic number increases by 1. Thus, the resulting nuclide is:\[\begin{align*}A_f &= A_i = 24\Z_f &= Z_i + 1 = 11 + 1 = 12\end{align*}\]The product is \(^{24}_{12}Mg\) (Magnesium-24).
05

Understanding Beta Plus Decay

In beta plus decay, a proton is converted into a neutron, and a positron is emitted. This decreases the atomic number by 1, while the mass number remains unchanged.
06

Beta Plus Decay of Oxygen-15

During the \(\beta^+\) decay of \(^{15}_{8}O\), the atomic number decreases by 1. Thus, the resulting nuclide is:\[\begin{align*}A_f &= A_i = 15\Z_f &= Z_i - 1 = 8 - 1 = 7\end{align*}\]The product is \(^{15}_{7}N\) (Nitrogen-15).

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Key Concepts

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

Alpha Decay
Radioactive decay is a fascinating process, and alpha decay is one important type. During alpha decay, an unstable nucleus releases an alpha particle. An alpha particle is similar to a helium nucleus, containing 2 protons and 2 neutrons. Let's consider the example of Plutonium-239. When it undergoes alpha decay, it emits an alpha particle, which reduces its mass number by 4 and its atomic number by 2. This means that after the decay, Plutonium-239 transforms into Uranium-235.
  • The original element is Plutonium (\(^{239}_{94}Pu\)).
  • After alpha decay, the element is Uranium (\(^{235}_{92}U\)).
It is crucial to remember that since alpha particles are quite massive, they do not penetrate materials deeply. Thus, even a piece of paper or the outer layer of human skin can stop them.
Beta Minus Decay
Beta minus decay is another common form of radioactive decay. In this process, a neutron within the nucleus transforms into a proton. Along with this transformation, a beta particle, which is a fast-moving electron, is also emitted. This increases the atomic number of the element by 1 but keeps its mass number unchanged.
Using the example from the textbook of Sodium-24, during beta minus decay, Sodium transforms into Magnesium-24:
  • The original nuclide is Sodium (\(^{24}_{11}Na\)).
  • After beta minus decay, we have Magnesium (\(^{24}_{12}Mg\)).
Beta particles are more penetrating than alpha particles but can still be blocked by materials like aluminum foil or a few meters of air.
Beta Plus Decay
The third type of radioactive decay is beta plus decay. This process involves the conversion of a proton into a neutron within the nucleus, with the emission of a positron (a particle with the same mass as an electron but a positive charge). This decay decreases the atomic number of the element by 1, but the mass number stays the same.
Consider the example of Oxygen-15. During beta plus decay, it changes into Nitrogen-15:
  • The original nuclide is Oxygen (\(^{15}_{8}O\)).
  • After beta plus decay, we have Nitrogen (\(^{15}_{7}N\)).
Positrons are similar to electrons in terms of penetration, and they typically collide with electrons, causing annihilation and releasing gamma radiation.

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Most popular questions from this chapter

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, \(^{205}_{81}Tl\)?

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?

The nucleus \(^{15}_{8}O\) has a half-life of 122.2 s; \(^{19}_{8}O\) has a half- life of 26.9 s. If at some time a sample contains equal amounts of \(^{15}_{8}O\) and \(^{19}_{8}O\), what is the ratio of \(^{15}_{8}O\) to \(^{19}_{8}O\) (a) after 3.0 min and (b) after 12.0 min?

A \(^6$$^0\)Co source with activity 2.6 \(\times\) 10\(^-$$^4\) Ci is embedded in a tumor that has mass 0.200 kg. The source emits \(\gamma\) photons with average energy 1.25 MeV. Half the photons are absorbed in the tumor, and half escape. (a) What energy is delivered to the tumor per second? (b) What absorbed dose (in rad) is delivered per second? (c) What equivalent dose (in rem) is delivered per second if the RBE for these g rays is 0.70? (d) What exposure time is required for an equivalent dose of 200 rem?

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?
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