Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![### Radioactive Decay Problem
**Instruction:**
Complete the following radioactive decay problem.
**Problem Statement:**
\[ \ce{^{210}_{84}Po -> ^{206}_{82}Pb +} \]
**Details:**
This equation represents a radioactive decay process where Polonium-210 (\(\ce{^{210}_{84}Po}\)) decays into Lead-206 (\(\ce{^{206}_{82}Pb}\)). The blank squares indicate missing particles that need to be determined to balance the nuclear equation.
### Explanation:
- **Polonium-210 (\(\ce{^{210}_{84}Po}\)):**
- Atomic number: 84
- Mass number: 210
- **Lead-206 (\(\ce{^{206}_{82}Pb}\)):**
- Atomic number: 82
- Mass number: 206
### Solution:
To balance the nuclear equation:
1. **Compare Mass Numbers:** \(210 = 206 + x\)
- Solving for \(x\): \(x = 210 - 206 = 4\)
2. **Compare Atomic Numbers:** \(84 = 82 + y\)
- Solving for \(y\): \(y = 84 - 82 = 2\)
Therefore, the particle emitted is:
\[ \ce{^{4}_{2}He} \] (Helium nucleus, also known as an alpha particle)
**Balanced Equation:**
\[ \ce{^{210}_{84}Po -> ^{206}_{82}Pb + ^{4}_{2}He} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f7b3938-790b-4169-a858-04369df79473%2Fc90c66f6-4451-4c0c-a045-c169fd5c2d2d%2Fygecdts_processed.png&w=3840&q=75)
Transcribed Image Text:### Radioactive Decay Problem
**Instruction:**
Complete the following radioactive decay problem.
**Problem Statement:**
\[ \ce{^{210}_{84}Po -> ^{206}_{82}Pb +} \]
**Details:**
This equation represents a radioactive decay process where Polonium-210 (\(\ce{^{210}_{84}Po}\)) decays into Lead-206 (\(\ce{^{206}_{82}Pb}\)). The blank squares indicate missing particles that need to be determined to balance the nuclear equation.
### Explanation:
- **Polonium-210 (\(\ce{^{210}_{84}Po}\)):**
- Atomic number: 84
- Mass number: 210
- **Lead-206 (\(\ce{^{206}_{82}Pb}\)):**
- Atomic number: 82
- Mass number: 206
### Solution:
To balance the nuclear equation:
1. **Compare Mass Numbers:** \(210 = 206 + x\)
- Solving for \(x\): \(x = 210 - 206 = 4\)
2. **Compare Atomic Numbers:** \(84 = 82 + y\)
- Solving for \(y\): \(y = 84 - 82 = 2\)
Therefore, the particle emitted is:
\[ \ce{^{4}_{2}He} \] (Helium nucleus, also known as an alpha particle)
**Balanced Equation:**
\[ \ce{^{210}_{84}Po -> ^{206}_{82}Pb + ^{4}_{2}He} \]
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