1. (1) What is the magnitude of the electric force of attraction between an iron nucleus (q = +26e) and its innermost electron if the distance between them is 1.5 × 10-¹2 m?
1. (1) What is the magnitude of the electric force of attraction between an iron nucleus (q = +26e) and its innermost electron if the distance between them is 1.5 × 10-¹2 m?
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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please type out your solution so that it is easy to read I have bad eyesight and cant read handriting well
![**Question 1:**
What is the magnitude of the electric force of attraction between an iron nucleus (q = +26e) and its innermost electron if the distance between them is 1.5 × 10⁻¹² m?
---
On an educational website, to facilitate understanding of this problem, the following steps can be provided to solve for the magnitude of the electric force:
**Step-by-Step Solution:**
1. **Identify the charge of the particles:**
- The charge of the iron nucleus (q) is \( +26e \), where \( e \) is the elementary charge (\( 1.602 \times 10^{-19} \, \text{C} \)).
- The charge of the electron is \( -e \).
2. **Determine the distance between the charges:**
- The given distance (r) between the iron nucleus and the innermost electron is \( 1.5 \times 10^{-12} \, \text{m} \).
3. **Use Coulomb’s Law to calculate the electric force:**
Coulomb’s Law formula:
\[
F = k_e \frac{|q_1 \cdot q_2|}{r^2}
\]
- \( F \) is the magnitude of the force between the charges.
- \( k_e \) is Coulomb's constant (\( 8.988 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)).
- \( q_1 \) and \( q_2 \) are the charges of the particles.
4. **Substitute the values into the equation:**
- Charge of the nucleus: \( q_1 = +26e = 26 \times 1.602 \times 10^{-19} \, \text{C} \)
- Charge of the electron: \( q_2 = -e = -1.602 \times 10^{-19} \, \text{C} \)
- Distance: \( r = 1.5 \times 10^{-12} \, \text{m} \)
The formula becomes:
\[
F = \left( 8.988 \times 10^9 \, \text{N} \cd](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F597e870a-d1dc-4e8b-ac25-38d259c0bf97%2Fea7b58de-f2ad-4c3b-8cb0-8073190866a7%2Fjbbla28_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 1:**
What is the magnitude of the electric force of attraction between an iron nucleus (q = +26e) and its innermost electron if the distance between them is 1.5 × 10⁻¹² m?
---
On an educational website, to facilitate understanding of this problem, the following steps can be provided to solve for the magnitude of the electric force:
**Step-by-Step Solution:**
1. **Identify the charge of the particles:**
- The charge of the iron nucleus (q) is \( +26e \), where \( e \) is the elementary charge (\( 1.602 \times 10^{-19} \, \text{C} \)).
- The charge of the electron is \( -e \).
2. **Determine the distance between the charges:**
- The given distance (r) between the iron nucleus and the innermost electron is \( 1.5 \times 10^{-12} \, \text{m} \).
3. **Use Coulomb’s Law to calculate the electric force:**
Coulomb’s Law formula:
\[
F = k_e \frac{|q_1 \cdot q_2|}{r^2}
\]
- \( F \) is the magnitude of the force between the charges.
- \( k_e \) is Coulomb's constant (\( 8.988 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)).
- \( q_1 \) and \( q_2 \) are the charges of the particles.
4. **Substitute the values into the equation:**
- Charge of the nucleus: \( q_1 = +26e = 26 \times 1.602 \times 10^{-19} \, \text{C} \)
- Charge of the electron: \( q_2 = -e = -1.602 \times 10^{-19} \, \text{C} \)
- Distance: \( r = 1.5 \times 10^{-12} \, \text{m} \)
The formula becomes:
\[
F = \left( 8.988 \times 10^9 \, \text{N} \cd
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