Chlorine has two stable nuclides, 35 Cl and 37 Cl. In contrast, 36 Cl is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of 36 Cl?

Welcome back. Everyone in this example, we're told that the three naturally occurring isotopes of potassium are potassium potassium 40 and potassium 41. The isotope, potassium 40 is radioactive and undergoes beta decay, identify the product of the radioactive decay of potassium In a beta decay we would begin with our isotope which in this prompt is going to be potassium 40. Recognize that on our periodic table potassium corresponds to atomic number and it's going to undergo a beta decay where we would have the product of a beta particle being our electron, which we would recall has a atomic number of negative one in a mass number of zero. And we would also produce a second unknown new glide product which we need to figure the identity of by balancing this equation. So we need to figure out its atomic number Z. And its mass number A. So figuring out for our mass number first. The mass number of our product. We're going to plug in our knowns based on the reactant side which is our potassium 40, we have a mass number of 40. This is set equal to our product side where we have a mass number of zero plus our product of our new glide mass number being a. Which is what we're solving for. And this means that we would say that a is going to equal 40. Now solving for our atomic number, which we'll do in purple will say that ZR atomic number of our product plugging in our knowns is equal to our atomic number of our potassium isotope, which is 19. Set equal to our product side. Where we have the atomic number of an electron particle being minus one plus are unknown atomic number being Z. And so solving for Z, we would have 19 plus one equal to Z. Where we would say that Z is equal to 20. And this means that now that we know our atomic number Z, recall that on our periodic table. This is going to correspond to the atom calcium. And so for our full Balanced nuclear equation, we would have our potassium which undergoes a beta decay to produce a beta electron particle And our product Calcium 40 with the atomic number 20. So for our final answer, this is going to be the product of our radioactive decay of of potassium 40. Sorry. So this is our final answer. I hope everything I explained was clear. If you have any questions, please leave them down below. And I will see everyone in the next practice video.

Ch.21 - Nuclear Chemistry

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Brown 15th Edition

Ch.21 - Nuclear Chemistry

Problem 74a

Chapter 21, Problem 74a

Chlorine has two stable nuclides, ³⁵Cl and ³⁷Cl. In contrast, ³⁶Cl is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of ³⁶Cl?

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Identify the type of decay: 36Cl undergoes beta decay, which involves the conversion of a neutron into a proton, emitting a beta particle (an electron) and an antineutrino.

Write the nuclear equation for beta decay: In beta decay, the atomic number increases by 1 while the mass number remains unchanged.

Determine the new element: Since the atomic number increases by 1, the element changes from chlorine (Cl) to the next element in the periodic table, which is argon (Ar).

Write the balanced nuclear equation: \( ^{36}_{17}\text{Cl} \rightarrow ^{36}_{18}\text{Ar} + \beta^- + \bar{\nu}_e \), where \( \beta^- \) is the beta particle and \( \bar{\nu}_e \) is the antineutrino.

Conclude the product: The product of the beta decay of 36Cl is 36Ar.

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Key Concepts

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

Beta Decay

Beta decay is a type of radioactive decay in which a beta particle (an electron or a positron) is emitted from an atomic nucleus. In the case of beta minus decay, a neutron is transformed into a proton, resulting in the emission of an electron and an antineutrino. This process increases the atomic number of the element by one, leading to the formation of a new element.

Nuclear Stability

Nuclear stability refers to the ability of an atomic nucleus to remain intact without undergoing radioactive decay. Stable nuclides, like 35Cl and 37Cl, have a balanced ratio of protons to neutrons, while unstable nuclides, such as 36Cl, have an imbalance that leads to decay. Understanding the factors that contribute to nuclear stability is essential for predicting the behavior of different isotopes.

Decay Products

Decay products are the new elements or isotopes formed as a result of radioactive decay. In the case of 36Cl undergoing beta decay, the decay product is 37Ar (argon), as the emission of a beta particle transforms the chlorine nucleus into an argon nucleus. Identifying decay products is crucial for understanding the implications of radioactive decay in various applications, including nuclear medicine and radiometric dating.

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Beta Decay

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The table provided gives the number of protons (p) and neutrons (n) for four isotopes, identified only as (i)–(iv). a. Write the symbol for each of the isotopes.

Textbook Question

Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

Textbook Question

Chlorine has two stable nuclides, ³⁵Cl and ³⁷Cl. In contrast, ³⁶Cl is a radioactive nuclide that decays by beta emission. (b) Based on the empirical rules about nuclear stability, explain why the nucleus of ³⁶Cl is less stable than either ³⁵Cl or ³⁷Cl.

Textbook Question

Nuclear scientists have synthesized approximately 1600 nuclei not known in nature. More might be discovered with heavy-ion bombardment using high-energy particle accelerators. Complete and balance the following reactions, which involve heavy-ion bombardments:

(a) ⁶₃Li + ⁵⁶₂₈Ni → ?

(b) ⁴⁰₂₀Ca + ²⁴⁸₉₆Cm → ¹⁴⁷₆₂Sm + ?

(d) ⁴⁰₂₀Ca + ²³⁸₉₂U → ⁷⁰₃₀Zn + 4 ¹₀n + 2 ?

Textbook Question

In 2010, a team of scientists from Russia and the United States reported creation of the first atom of element 117, which is named tennessine, and whose symbol is Ts. The synthesis involved the collision of a target of ²⁴⁹₉₇Bk with accelerated ions of an isotope which we will denote Q. The product atom, which we will call Z, immediately releases neutrons and forms ²⁹⁴₁₁₇Ts: ²⁴⁹₉₇Bk + Q → Z → ²⁹⁴₁₁₇Ts + 3 ¹₀n (a) What are the identities of isotopes Q and Z? (c) Collision of ions of isotope Q with a target was also used to produce the first atoms of livermorium, Lv. The initial product of this collision was ²⁹⁶₁₁₆Lv. What was the target isotope with which Q collided in this experiment?

Chlorine has two stable nuclides, 35Cl and 37Cl. In contrast, 36Cl is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of 36Cl?

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Key Concepts

Beta Decay

Nuclear Stability

Decay Products

Chlorine has two stable nuclides, ³⁵Cl and ³⁷Cl. In contrast, ³⁶Cl is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of ³⁶Cl?