Calculate the energy in eV for the ground state and first two excited states of an electron in an infinite potential well of width 1 Å.

To calculate the energy levels of an electron in an infinite potential well (also known as a particle in a box), we use the following formula for the energy levels \( E_n \):

\[ E_n = \frac{n^2 h^2}{8mL^2} \]

where:
- \( n \) is the quantum number (1 for ground state, 2 for first excited state, etc.),
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ Js} \)),
- \( m \) is the mass of the electron (\( 9.109 \times 10^{-31} \text{ kg} \)),
- \( L \) is the width of the potential well.

Given:
- Width of the well \( L = 1 \text{ Å} = 1 \times 10^{-10} \text{ m} \)

Let's calculate the energies for \( n = 1 \) (ground state), \( n = 2 \) (first excited state), and \( n = 3 \) (second excited state).

### Ground State (\( n = 1 \)):
\[ E_1 = \frac{(1)^2 \cdot (6.626 \times 10^{-34})^2}{8 \cdot 9.109 \times 10^{-31} \cdot (1 \times 10^{-10})^2} \]
\[ E_1 = \frac{6.626^2 \times 10^{-68}}{8 \cdot 9.109 \times 10^{-31} \cdot 10^{-20}} \]
\[ E_1 = \frac{43.91 \times 10^{-68}}{7.287 \times 10^{-50}} \]
\[ E_1 = 6.026 \times 10^{-19} \text{ J} \]

Convert \( E_1 \) from joules to electron volts (eV):
\[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]
\[ E_1 = \frac{6.026 \times 10^{-19}}{1.602 \times 10^{-19}} \text{ eV} \]
\[ E_1 \approx 3.76 \text{ eV} \]

### First Excited State (\( n = 2 \)):
\[ E_2 = \frac{(2)^2 \cdot (6.626 \times 10^{-34})^2}{8 \cdot 9.109 \times 10^{-31} \cdot (1 \times 10^{-10})^2} \]
\[ E_2 = 4 \cdot E_1 \]
\[ E_2 = 4 \cdot 3.76 \text{ eV} \]
\[ E_2 \approx 15.04 \text{ eV} \]

### Second Excited State (\( n = 3 \)):
\[ E_3 = \frac{(3)^2 \cdot (6.626 \times 10^{-34})^2}{8 \cdot 9.109 \times 10^{-31} \cdot (1 \times 10^{-10})^2} \]
\[ E_3 = 9 \cdot E_1 \]
\[ E_3 = 9 \cdot 3.76 \text{ eV} \]
\[ E_3 \approx 33.84 \text{ eV} \]

### Summary of Energies:
- Ground state energy (\( E_1 \)): \( \approx 3.76 \text{ eV} \)
- First excited state energy (\( E_2 \)): \( \approx 15.04 \text{ eV} \)
- Second excited state energy (\( E_3 \)): \( \approx 33.84 \text{ eV} \)

Do you want details or have any questions?

Here are some related questions:
1. How does the width of the potential well affect the energy levels?
2. What would the energy levels be if the width of the well were doubled?
3. How would the energy levels change if the particle were a proton instead of an electron?
4. What is the significance of quantized energy levels in quantum mechanics?
5. Can you explain the concept of a potential well in more detail?
6. How does the infinite potential well model apply to real-world systems?
7. What is the physical meaning of the ground state and excited states?
8. How are energy levels determined for a finite potential well?

**Tip:** Understanding the relationship between the width of the potential well and the energy levels helps in grasping the concepts of quantum confinement and its applications in nanotechnology.

Learn how to calculate the energy levels of an electron in an infinite potential well with a width of 1 Å. This includes the ground state and first two excited states, calculated using quantum mechanics principles and the energy levels formula. Ground state energy is approximately 3.76 eV, first excited state is approximately 15.04 eV, and second excited state is approximately 33.84 eV.

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