Fill in the missing particles in each nuclear equation. d. 75 35 Br → ____ + 0 +1 e

Welcome back everyone in this example. We need to give the bounced nuclear equation for the positron emission of roaming 75. So let's go ahead and begin with our target new glide which is given in the prompt as bromine, 75, being its mass number which we recall is represented in the left hand exponents. When we refer to our periodic tables to find bro mean we see that it corresponds to atomic number 35 found in the D. Block of our periodic table. And we know that this is going to be producing a positron particle, which we should recall is represented by the symbol E. And has a atomic number of plus one and a mass number of zero. And we need to figure out the unknown product new glide produced from our positron emission of Rome in 75. So we are going to identify this unknown product X. Where we need to figure out its mass number a and its atomic number Z. So we're going to have to set up an expression with our knowns to sulfur are unknowns A and C. Beginning with finding a. Let's plug in what we know we know our mass number of romaine being 75 which is set equal to our mass number of our positron particle being zero and added to our mass number of our unknown product. New glide being A. And so solving for A. We would say that A. And let's use the color red a. is equal to 75. Moving forward. In our solution, we want to figure out the atomic number of our unknown product new glide. So plugging in our notes, we begin with the atomic number of bromine which we know is equal to our products. Where we have our atomic number of our positron particle being one which is added to our unknown atomic number of our product, new Clyde being Z. So solving for Z, we would say that Z is equal to 35 minus one and so Z is equal to 34. And so we would refer to our periodic table and see that on our periodic table. Atomic number 34 corresponds to the atoms selenium. And so now we can write out our full bounce nuclear equation where we have roaming 75 with with its atomic number 35 which undergoes a positron emission to produce a positron particle with the mass number of zero and atomic number plus one. And our product new glide, which would be our selenium with the atomic number 34 mass number 75. And our final answer is going to be this entire balanced nuclear equation that we've come up with. So I hope everything I explained was clear. If you have any questions, leave them down below and I will see everyone in the next practice video

Ch.21 - Radioactivity & Nuclear Chemistry

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Ch.21 - Radioactivity & Nuclear Chemistry

Problem 36d

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Chapter 21, Problem 36d

Fill in the missing particles in each nuclear equation. d. ⁷⁵₃₅Br → ____ + ⁰₊₁e

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Identify the type of decay occurring in the given nuclear equation. The presence of 0+1e, which represents a positron, indicates that the decay process is positron emission.

Understand that in positron emission, a proton in the nucleus is converted into a neutron, and a positron is emitted. This process decreases the atomic number by one but keeps the mass number the same.

Write the general form of the nuclear equation for positron emission: AZX → AZ-1Y + 0+1e, where A is the mass number, Z is the atomic number, X is the parent nucleus, and Y is the daughter nucleus.

Apply the general form to the specific case: 7535Br → 7534Y + 0+1e. Here, the atomic number decreases by one, so the element symbol changes from Br (bromine) to Se (selenium), which has an atomic number of 34.

Complete the nuclear equation by filling in the missing particle: 7535Br → 7534Se + 0+1e.

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Key Concepts

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

Nuclear Reactions

Nuclear reactions involve changes in an atom's nucleus, resulting in the transformation of one element into another. These reactions can include processes such as alpha decay, beta decay, and gamma emission. Understanding the type of nuclear reaction is crucial for predicting the products formed, as different reactions follow specific rules regarding particle conservation and transformation.

Beta Decay

Beta decay is a type of radioactive decay in which a neutron in the nucleus is transformed into a proton, emitting a beta particle (an electron or positron) in the process. This transformation increases the atomic number of the element by one while keeping the mass number unchanged. Recognizing beta decay is essential for completing nuclear equations, as it directly influences the identity of the resulting nuclide.

Conservation of Nucleons

In nuclear reactions, the conservation of nucleons (protons and neutrons) must be maintained. This principle states that the total number of nucleons before and after the reaction must remain constant. When filling in missing particles in a nuclear equation, it is important to ensure that the sum of the atomic numbers and mass numbers on both sides of the equation is equal, reflecting this conservation law.