Calculate energy of the first three quantum states for an electron in a 1D potential well of width 0.1 nm
Answer:-
Formula:
E_n = \frac{n^2 \pi^2 \hbar^2}{2 m a^2} <br />Where:
- n = 1, 2, 3 \ldots (quantum number)
- \hbar = \frac{h}{2\pi} = 1.055 \times 10^{-34} , \text{J·s}
- m = 9.1 \times 10^{-31} , \text{kg} (mass of electron)
- a = 0.1 , \text{nm} = 0.1 \times 10^{-9} , \text{m} = 1 \times 10^{-10} , \text{m}
Energy of the First Three Quantum States in a 1D Infinite Potential Well
Given:
- Width of well (a) = 0.1 nm = 1 × 10−10 m
- Mass of electron (m) = 9.1 × 10−31 kg
- ℏ = h / 2π = 1.055 × 10−34 J·s
Formula:
Eₙ = (n² × π² × ℏ²) / (2 × m × a²)
Energy Calculations:
For n = 1:
E₁ = (1² × π² × (1.055 × 10−34)²) / (2 × 9.1 × 10−31 × (1 × 10−10)²)
≈ 6.02 × 10−18 J ≈ 37.6 eV
For n = 2:
E₂ = 4 × E₁ ≈ 150.4 eV
For n = 3:
E₃ = 9 × E₁ ≈ 338.4 eV
Final Answers:
- First state (n = 1): 37.6 eV
- Second state (n = 2): 150.4 eV
- Third state (n = 3): 338.4 eV