What nuclide is produced in the following radioactive decays? (a) α \alpha decay of 94 239 P u _{94}^{239}Pu (b) β − \beta^{-} decay of 11 24 N a _{11}^{24}Na (c) β + \beta^{+} decay of 8 15 O _8^{15}O

Hi, everyone. In this practice problem, we're being asked to determine the daughter isotope that is going to be produced by the following reaction. First negative beta decay of phosphorus, 32 15 p, second alpha decay of radium 226 88 radium and third positive beta decay of carbon, 11 6 carbon. The options listed the different daughter isotopes for different decay equations. So in this practice problem, the way we want to tackle this problem is to remember that the left side and the right side of the A and the C or the total mass on each side of the decay equation, the total charge and the nucleon number should be conserved. So in this case, let A Z X B, the daughter isotope for each decay equation. And let's start by actually tackling the first part which is going to be the negative decay or negative beta decay of phosphorus. 32 15 B. For an uh negative beta decay, we have an emission of an electron, the total charge and the nucleon number should be conserved. So 32 15 P will decay into zero minus one E plus AC X So on the left side, we have 15, which will also uh be, well, which will then be equal to, on the right side minus one plus C for the Z. And for the A, on the left side, we have 32 which will be equal to zero plus A. So with this two equation, we have the Z to then be equal to 16 and the A to then be equals to 32. And then we can notice that the atomic number of the daughter nude increases by one while its atomic mass remains constant. So then with the uh the A of 32 the Z of 16, we have the daughter nucleus to be 16 sulfur or s So that will be the answer for the first part. And then moving on to the second part, we have the alpha decay of radium, which will have the uh decay equation of 226 radium decay into a Z X plus four to helium because it is an alpha decay. So looking into the Z, we have 88 on the left side, which will be equals to Z plus two. So then our Z for the daughter isotope will then be equal to 86 of uh looking at the A, on the left side, we have 226 A plus four. And then that will give us an A of 22 on the daughter isotope. So the total charge and the nuclear number are conserved through this ac uh values. And the atomic number of the daughter nude decreases by two and its, and its atomic mass decreases by four. So according to this, we can uh determine that the daughter nucleus is going to be redden, which is going to be 222 86. Er And lastly looking into the third equation or the third part of our problem, which will be the uh positive beta decay beta plus of carbon 11. For this decay, we will have an emission of positron. So the decay equation for carbon 11 will be 11 6 C or carbon 11 decaying into zero plus one E or positron plus AC X. Looking at the Zs, we have six on the left side plus one on the right side plus C. So then our Z will then uh Z $40 isotope will then be equals to five. And then looking at the ace, we have 11 on the left side, zero plus A and the A will still remains to be 11 of on the daughter aceto. So the total charge and the nuclear number are conserved through this and the atomic number of the daughter newly decreased by one while its atomic mass remains constant. Therefore, we can determine that the daughter nucleus is going to be 5 boron or 11 5 B. So there we have it for this practice problem. The first uh decay equation will have the daughter isotope 32 16 S or sulfur. And then for the second one, it will have 222 86 Redden or R N. And for the last part, it will have 11 5 boron or B. So all of this will correspond to option A in our answer choices. So option A will be the answer to this particular practice problem. So that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.

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Problem 13

Textbook Question

What nuclide is produced in the following radioactive decays?
(a) $α$ decay of $sub 94 to the 239th power P u$
(b) $β^{-}$ decay of $sub 11 to the 24th power N a$
(c) $β^{+}$ decay of $sub 8 to the 15th power O$

Verified step by step guidance

Step 1: Understand the type of decay and its effect on the nucleus. In alpha (α) decay, the nucleus emits an alpha particle (2 protons and 2 neutrons), reducing the atomic number by 2 and the mass number by 4. In beta-minus (β⁻) decay, a neutron converts into a proton, increasing the atomic number by 1 while the mass number remains unchanged. In beta-plus (β⁺) decay, a proton converts into a neutron, decreasing the atomic number by 1 while the mass number remains unchanged.

Step 2: For part (a), identify the parent nuclide as ²³⁹₉₄Pu (Plutonium-239). Apply the rules of alpha decay: subtract 2 from the atomic number (94 - 2 = 92) and subtract 4 from the mass number (239 - 4 = 235). The resulting nuclide is ²³⁵₉₂U (Uranium-235).

Step 3: For part (b), identify the parent nuclide as ²⁴₁₁Na (Sodium-24). Apply the rules of beta-minus decay: increase the atomic number by 1 (11 + 1 = 12) while keeping the mass number unchanged (24). The resulting nuclide is ²⁴₁₂Mg (Magnesium-24).

Step 4: For part (c), identify the parent nuclide as ¹⁵₈O (Oxygen-15). Apply the rules of beta-plus decay: decrease the atomic number by 1 (8 - 1 = 7) while keeping the mass number unchanged (15). The resulting nuclide is ¹⁵₇N (Nitrogen-15).

Step 5: Summarize the results: (a) ²³⁵₉₂U is produced from alpha decay of ²³⁹₉₄Pu, (b) ²⁴₁₂Mg is produced from beta-minus decay of ²⁴₁₁Na, and (c) ¹⁵₇N is produced from beta-plus decay of ¹⁵₈O.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This can occur in various forms, including alpha decay, beta decay (both beta-minus and beta-plus), and gamma decay. Each type of decay results in the transformation of the original nuclide into a different nuclide, often accompanied by the release of particles or electromagnetic radiation.

Alpha Decay

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, consisting of two protons and two neutrons. This process decreases the atomic number by two and the mass number by four, resulting in a new element. For example, when plutonium-239 undergoes alpha decay, it transforms into uranium-235.

Beta Decay

Beta decay is a process where a neutron in an atomic nucleus is transformed into a proton, emitting a beta particle (an electron or positron) in the process. In beta-minus decay, a neutron converts to a proton and emits an electron, increasing the atomic number by one. Conversely, in beta-plus decay, a proton converts to a neutron, emitting a positron and decreasing the atomic number by one, leading to the formation of a different nuclide.

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