6.8: A 6 eV electron is confined in an infinite 1 D potential well. The region between the potential wall is 1.00 nm long. Calculate the electrons quantum number in its current state.
6.8: A 6 eV electron is confined in an infinite 1 D potential well. The region between the potential wall is 1.00 nm long. Calculate the electrons quantum number in its current state.
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![**Problem 6.8: Calculating the Quantum Number of an Electron in an Infinite 1D Potential Well**
**Statement:**
A 6 eV electron is confined in an infinite 1D potential well. The region between the potential walls is 1.00 nm long. Calculate the electron's quantum number in its current state.
**Explanation:**
In quantum mechanics, an electron confined in an infinite potential well (also known as particle in a box) can only occupy certain energy levels, which are quantized. The energy levels for an electron in a 1D potential well of width \( L \) are given by the equation:
\[ E_n = \dfrac{n^2 h^2}{8mL^2} \]
Where:
- \( E_n \) is the energy of the electron at quantum number \( n \)
- \( n \) is the quantum number (a positive integer)
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) Js)
- \( m \) is the mass of the electron (\( 9.109 \times 10^{-31} \) kg)
- \( L \) is the width of the potential well
Given that:
- \( E_n = 6 \) eV (1 eV = \( 1.602 \times 10^{-19} \) J)
- \( L = 1.00 \) nm (\( 1 \) nm = \( 1 \times 10^{-9} \) m)
**Calculation:**
1. Convert the given energy into joules:
\[ E_n = 6 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 9.612 \times 10^{-19} \, \text{J} \]
2. Rearrange the energy equation to solve for \( n \):
\[ n^2 = \dfrac{8mL^2E_n}{h^2} \]
3. Plug in the values to compute \( n \):
\[ n^2 = \dfrac{8 \times 9.109 \times 10^{-31} \times (1.00 \times 10^{-9})^2 \times 9.612 \times 10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fe8677b-d2b4-4cf1-b1ed-08820154fcb5%2F1882fdf0-af06-4c4f-8f84-4af19f014a4f%2Fy5mp4zf_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 6.8: Calculating the Quantum Number of an Electron in an Infinite 1D Potential Well**
**Statement:**
A 6 eV electron is confined in an infinite 1D potential well. The region between the potential walls is 1.00 nm long. Calculate the electron's quantum number in its current state.
**Explanation:**
In quantum mechanics, an electron confined in an infinite potential well (also known as particle in a box) can only occupy certain energy levels, which are quantized. The energy levels for an electron in a 1D potential well of width \( L \) are given by the equation:
\[ E_n = \dfrac{n^2 h^2}{8mL^2} \]
Where:
- \( E_n \) is the energy of the electron at quantum number \( n \)
- \( n \) is the quantum number (a positive integer)
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) Js)
- \( m \) is the mass of the electron (\( 9.109 \times 10^{-31} \) kg)
- \( L \) is the width of the potential well
Given that:
- \( E_n = 6 \) eV (1 eV = \( 1.602 \times 10^{-19} \) J)
- \( L = 1.00 \) nm (\( 1 \) nm = \( 1 \times 10^{-9} \) m)
**Calculation:**
1. Convert the given energy into joules:
\[ E_n = 6 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 9.612 \times 10^{-19} \, \text{J} \]
2. Rearrange the energy equation to solve for \( n \):
\[ n^2 = \dfrac{8mL^2E_n}{h^2} \]
3. Plug in the values to compute \( n \):
\[ n^2 = \dfrac{8 \times 9.109 \times 10^{-31} \times (1.00 \times 10^{-9})^2 \times 9.612 \times 10
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