The nucleus of a 125 Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm i diameter. It has 54 protons and charge q= +54e (1 fm = 1 femtometer = 1×10-15m.) Hint: Treat the field of the spherical nucleus as that of a point charge at the center. Part A What is the strength of the electric field 7.0 fm from the surface of the nucleus? Part B What is the direction of the electric field at the point?
The nucleus of a 125 Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm i diameter. It has 54 protons and charge q= +54e (1 fm = 1 femtometer = 1×10-15m.) Hint: Treat the field of the spherical nucleus as that of a point charge at the center. Part A What is the strength of the electric field 7.0 fm from the surface of the nucleus? Part B What is the direction of the electric field at the point?
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
Transcribed Image Text:### Properties of the Xe-125 Nucleus and Electric Field Calculation
The nucleus of a \( ^{125}\text{Xe} \) atom (an isotope of the element xenon with a mass of 125 u) has a diameter of 6.0 femtometers (fm). The nucleus contains 54 protons, giving it a total charge of \( q = +54e \). Here, \( e \) denotes the elementary charge, and \( 1 \, \text{fm} = 1 \times 10^{-15} \, \text{meters} \).
The electric field around this nucleus can be approximated as if the charge were a point located at the center of the spherical nucleus.
#### Part A
**Question:** What is the strength of the electric field at a distance of 7.0 fm from the surface of the nucleus?
#### Part B
**Question:** What is the direction of the electric field at this point?
### Explanation:
- **Diagram and Calculation Context:**
- The nucleus is to be considered as a spherical point charge centered at a specific point.
- Distance calculations involve finding the electric field just outside this point charge.
- **Graph/Diagram Description:**
- Imagine a spherical nucleus with a central point charge inside it.
- A point located 7.0 fm from the surface would be drawn such that the total distance from the center of the nucleus to this point is \( (6/2 + 7) \) fm (sum of the nucleus radius and the extra distance from the surface).
This exercise aims to teach high school or first-year university students about electrostatic fields around atomic nuclei by considering an approximation that simplifies the math typically used in advanced physics.
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