Identify the unknown isotope X X in the following decays. X → 40 K + e + + ν \text{X}\rightarrow^{40}\text{K}+\text{e}^{+}+\nu

Welcome back. Everyone in this problem, an unknown isotope y decay equation is Y decays to titanium 50 plus opposition plus a neutrino. And we want to identify the unknown isotope Y in the decay. For our answer choices. A is oxygen 18 B is chlorine 37 C is argo 40 D is vanadium 50. Now, what's going on here? We want to figure out our unknown isotope. Why if we look at our decay equation, notice that it's emitting a positron and a neutrino. So this represents a beta plus decay because in a beta decay, a proton in the nucleus is converted to a neutron and emits a positron and a neutrino. Now, two things happen in a beta plus decay when it comes to the atomic number. One is that in a beta decay, it results in a decrease of one unit. OK, in the atomic number, right? But there's no change in the atomic mass, no change in the atomic mass of the nucleus. OK. So if we think backwards, we should be able to figure out what the unknown isotope is. Let's use our first idea. It decreases by one unit. Now, if we look at titanium 50. OK. The atomic number for titanium 50 is 22. OK. So that means the atomic number for the unknown isotope must be 23. So the atomic number where is a top or for why must be equal to 23? Because it would have been 23 minus one that gives us 22. So do we know what element has an atomic number of 23? It must be vanadium. So that's one way. And you know, the unknown isotope has to be vanadium. Let's use our second parameter. There's no change in the atomic mass. So because there's no change in the atomic mass, this also tells us that the atomic mass of Y is also going to be equal to 50. And if we look in our answer choices, the only choice with an atomic mass of 50 is vanadium 50. So since it meets both criteria, then our unknown isotope is vanadium 50. Thanks a lot for watching everyone. I hope this video helped.

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Problem 27c

Textbook Question

Identify the unknown isotope $X$ in the following decays. $X \to^{40} K + e^{+} + ν$

Verified step by step guidance

Understand the problem: The given decay involves an unknown isotope (X) decaying into potassium-40 (40K), a positron (e+), and a neutrino (ν). This is a beta-plus (β+) decay, where a proton in the nucleus of the parent isotope is converted into a neutron, emitting a positron and a neutrino.

Apply the principle of conservation of nucleon number: The total number of nucleons (protons + neutrons) must remain the same before and after the decay. Since 40K has a mass number of 40, the unknown isotope X must also have a mass number of 40.

Apply the principle of conservation of charge: In β+ decay, the atomic number of the daughter nucleus decreases by 1 because a proton is converted into a neutron. Potassium (K) has an atomic number of 19, so the unknown isotope X must have an atomic number of 20 (19 + 1).

Identify the element with atomic number 20: Using the periodic table, the element with atomic number 20 is calcium (Ca). Therefore, the unknown isotope is calcium-40 (40Ca).

Summarize the result: The unknown isotope X is identified as calcium-40 (40Ca), which undergoes β+ decay to produce potassium-40 (40K), a positron, and a neutrino.

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Key Concepts

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

Isotopes

Isotopes are variants of a particular chemical element that have the same number of protons but different numbers of neutrons. This results in different atomic masses for the isotopes of the same element. Understanding isotopes is crucial in nuclear physics, as they can undergo various types of decay, affecting their stability and the reactions they participate in.

Beta Plus Decay

Beta plus decay is a type of radioactive decay in which a proton in the nucleus of an atom is transformed into a neutron, emitting a positron (e+) and a neutrino (ν) in the process. This decay decreases the atomic number of the element by one while keeping the mass number constant. Recognizing this decay process is essential for identifying the parent isotope and the resulting daughter isotope.

Conservation of Energy and Momentum

In nuclear reactions, the conservation of energy and momentum principles state that the total energy and momentum before the decay must equal the total energy and momentum after the decay. This principle helps in analyzing decay processes and determining the properties of the emitted particles, such as the positron and neutrino in beta plus decay, ensuring that the reaction adheres to fundamental physical laws.

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