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To solve this problem, we need to calculate the energy difference between the ground state and the first excited state of an electron in a one-dimensional box (particle in a box model). Here’s how we can approach it:

### Given Data
- **Length of the box, \( L = 10^{-8} \, \text{cm} = 10^{-10} \, \text{m} \)**
- **Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{J}\cdot\text{s} \)**
- **Mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)**

### Formula for Energy Levels in a 1D Box
For a particle in a one-dimensional box, the energy levels are given by:
\[
E_n = \frac{n^2 h^2}{8 m L^2}
\]
where:
- \( E_n \) is the energy of the \( n \)-th level,
- \( n \) is the quantum number (1 for the ground state, 2 for the first excited state, etc.),
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle (electron in this case),
- \( L \) is the length of the box.

### Step 1: Calculate \( E_1 \) (Ground State Energy)
For the ground state (\( n = 1 \)):
\[
E_1 = \frac{1^2 \cdot h^2}{8 m L^2} = \frac{h^2}{8 m L^2}
\]

### Step 2: Calculate \( E_2 \) (First Excited State Energy)
For the first excited state (\( n = 2 \)):
\[
E_2 = \frac{2^2 \cdot h^2}{8 m L^2} = \frac{4 h^2}{8 m L^2} = \frac{h^2}{2 m L^2}
\]

### Step 3: Find the Energy Difference \( \Delta E = E_2 - E_1 \)
\[
\Delta E = E_2 - E_1 = \frac{h^2}{2 m L^2} - \frac{h^2}{8 m L^2} = \frac{4 h^2 - h^2}{8 m L^2} = \frac{3 h^2}{8 m L^2}
\]

### Substituting Values
Now we can plug in the given values for \( h \), \( m \), and \( L \):
\[
\Delta E = \frac{3 \times (6.63 \times 10^{-34})^2}{8 \times (9.1 \times 10^{-31}) \times (10^{-10})^2}
\]

### Calculating the Result
1. Compute \( h^2 = (6.63 \times 10^{-34})^2 \).
2. Compute \( m \times L^2 = 9.1 \times 10^{-31} \times (10^{-10})^2 \).
3. Divide to find \( \Delta E \).

I will carry out these calculations now.The energy difference (\( \Delta E \)) between the ground state and the first excited state of the electron is approximately:
\[
\Delta E \approx 1.81 \times 10^{-17} \, \text{J}
\]

Would you like a detailed breakdown of each calculation step or have any other questions?

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### Related Questions:
1. How would the energy difference change if the length of the box were doubled?
2. What is the energy difference between the first and second excited states?
3. How does the particle-in-a-box model apply to other particles, like protons or neutrons?
4. What impact does an increase in electron mass have on the energy levels?
5. Can we use this model for particles in a two- or three-dimensional box?

### Tip:
For quantum mechanical problems, remember that as the box length \( L \) decreases, the energy levels become more widely spaced due to the \( L^2 \) term in the denominator.

Calculate the energy difference between the ground state and first excited state for electron if the length of the box is 10^-8 cm. (h = 6.63 x 10^-34 J-sec, m = 9.1 x 10^-31 kg).

Learn how to calculate the energy difference between the ground state and first excited state of an electron confined in a 10^-8 cm box. This problem uses quantum mechanics principles, such as the particle in a box model and energy level calculations. Suitable for college physics students.

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