QUESTION 5 Find the lowest three energies in eV for an electron in a one dimensional box of length about the size of an atom
Question
Basic Answer
Step 1: Define the problem
We need to find the three lowest energy levels for an electron confined to a one-dimensional box. We’ll assume the length of the box (L) is approximately the size of a hydrogen atom, which is roughly 1 Å (1 x 10⁻¹⁰ m). We’ll use the particle in a box model.
Step 2: Relevant Equation
The energy levels for a particle in a 1D box are given by:
En = (n²h²)/(8mL²)
where:
- n = quantum number (n = 1, 2, 3…)
- h = Planck’s constant (6.626 x 10⁻³⁴ Js)
- m = mass of electron (9.109 x 10⁻³¹ kg)
- L = length of the box (1 x 10⁻¹⁰ m)
Step 3: Calculate the lowest three energy levels
We’ll calculate E1, E2, and E3 by substituting n = 1, 2, and 3 respectively into the equation. Note that the result will be in Joules, so we’ll need to convert to electron volts (eV) using the conversion factor 1 eV = 1.602 x 10⁻¹⁹ J.
E1 = (1² * (6.626 x 10⁻³⁴)²)/(8 * 9.109 x 10⁻³¹ * (1 x 10⁻¹⁰)²) ≈ 6.02 x 10⁻¹⁸ J ≈ 37.6 eV
E2 = (2² * (6.626 x 10⁻³⁴)²)/(8 * 9.109 x 10⁻³¹ * (1 x 10⁻¹⁰)²) ≈ 2.41 x 10⁻¹⁷ J ≈ 150.3 eV
E3 = (3² * (6.626 x 10⁻³⁴)²)/(8 * 9.109 x 10⁻³¹ * (1 x 10⁻¹⁰)²) ≈ 5.42 x 10⁻¹⁷ J ≈ 338.5 eV
Final Answer
The three lowest energies for an electron in a 1 Å box are approximately 37.6 eV, 150.3 eV, and 338.5 eV. Note that these values are approximate due to rounding. The actual size of an atom can vary, leading to slightly different energy levels.
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