College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Nuclear Reaction Completion: Educational Guide**
In this example, you're asked to determine which element completes the nuclear reaction:
\[
\frac{2}{1} \text{H} + \frac{98}{42} \text{Mo} \rightarrow \, ? + \frac{1}{0} \text{n}
\]
The options provided are:
- \( \frac{97}{42} \text{Mo} \)
- \( \frac{100}{44} \text{Ru} \)
- \( \frac{93}{41} \text{Nb} \)
- \( \frac{99}{43} \text{Tc} \)
**Explanation:**
In nuclear reactions, the sum of atomic numbers (the subscript) and the mass numbers (the superscript) must remain constant on both sides of the equation to satisfy the law of conservation of mass and charge.
- Atomic numbers on the left: \(1 (\text{H}) + 42 (\text{Mo}) = 43\)
- Possible atomic numbers after neutron emission: 43
- Mass numbers on the left: \(2 (\text{H}) + 98 (\text{Mo}) = 100\)
- Possible mass numbers after neutron emission: 99 (since one neutron, with mass number 1, is emitted)
Using these criteria, the option \( \frac{99}{43} \text{Tc} \) completes the reaction:
\[
\frac{2}{1} \text{H} + \frac{98}{42} \text{Mo} \rightarrow \frac{99}{43} \text{Tc} + \frac{1}{0} \text{n}
\]
This educational example illustrates how conservation laws apply in nuclear reactions, helping students understand isotopic changes during nuclear transformations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F691a14d9-61ac-439d-ab51-0a05f8ab7f1d%2F1179176d-a090-4b0f-b552-9a072ae51884%2Fub3wzo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Nuclear Reaction Completion: Educational Guide**
In this example, you're asked to determine which element completes the nuclear reaction:
\[
\frac{2}{1} \text{H} + \frac{98}{42} \text{Mo} \rightarrow \, ? + \frac{1}{0} \text{n}
\]
The options provided are:
- \( \frac{97}{42} \text{Mo} \)
- \( \frac{100}{44} \text{Ru} \)
- \( \frac{93}{41} \text{Nb} \)
- \( \frac{99}{43} \text{Tc} \)
**Explanation:**
In nuclear reactions, the sum of atomic numbers (the subscript) and the mass numbers (the superscript) must remain constant on both sides of the equation to satisfy the law of conservation of mass and charge.
- Atomic numbers on the left: \(1 (\text{H}) + 42 (\text{Mo}) = 43\)
- Possible atomic numbers after neutron emission: 43
- Mass numbers on the left: \(2 (\text{H}) + 98 (\text{Mo}) = 100\)
- Possible mass numbers after neutron emission: 99 (since one neutron, with mass number 1, is emitted)
Using these criteria, the option \( \frac{99}{43} \text{Tc} \) completes the reaction:
\[
\frac{2}{1} \text{H} + \frac{98}{42} \text{Mo} \rightarrow \frac{99}{43} \text{Tc} + \frac{1}{0} \text{n}
\]
This educational example illustrates how conservation laws apply in nuclear reactions, helping students understand isotopic changes during nuclear transformations.
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