Butadiene contains 4π electrons, each of which moves freely in the molecule. Assume the molecule in a one-dimension box. The energy required for the transition of an electron to th... Butadiene contains 4π electrons, each of which moves freely in the molecule. Assume the molecule in a one-dimension box. The energy required for the transition of an electron to the first excited state will be ΔE = ?
Understand the Problem
The question is asking to calculate the energy transition (ΔE) of an electron moving from the ground state to the first excited state in a one-dimensional box model for butadiene, which contains 4 π electrons. To solve this, we will use the quantum mechanical model of a particle in a box.
Answer
$$ ΔE = \frac{3h^2}{8m L^2} $$
Answer for screen readers
The energy transition (ΔE) for the electron moving from the ground state to the first excited state is given by:
$$ ΔE = \frac{3h^2}{8m L^2} $$
Steps to Solve
- Identify the Quantum States For a one-dimensional box, the energy levels are given by the formula:
$$ E_n = \frac{n^2 h^2}{8 m L^2} $$
where:
- (E_n) is the energy of the (n^{th}) state,
- (n) is the principal quantum number (1 for ground state, 2 for first excited state),
- (h) is Planck's constant,
- (m) is the mass of the electron,
- (L) is the length of the box.
- Calculate Energy of the Ground State (n=1) Substituting (n = 1) into the energy formula:
$$ E_1 = \frac{1^2 h^2}{8 m L^2} = \frac{h^2}{8 m L^2} $$
- Calculate Energy of the First Excited State (n=2) Now, substituting (n = 2):
$$ E_2 = \frac{2^2 h^2}{8 m L^2} = \frac{4 h^2}{8 m L^2} = \frac{h^2}{2 m L^2} $$
- Determine the Energy Transition (ΔE) The transition energy, (ΔE), from the ground state to the first excited state is found by subtracting the ground state energy from the excited state energy:
$$ ΔE = E_2 - E_1 $$
Substituting the previous results:
$$ ΔE = \frac{h^2}{2 m L^2} - \frac{h^2}{8 m L^2} $$
- Simplify the Expression To simplify, factor out the common terms:
$$ ΔE = \left( \frac{h^2}{m L^2} \right) \left( \frac{1}{2} - \frac{1}{8} \right) $$
Calculating the fraction:
$$ \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8} $$
Thus:
$$ ΔE = \frac{3h^2}{8m L^2} $$
The energy transition (ΔE) for the electron moving from the ground state to the first excited state is given by:
$$ ΔE = \frac{3h^2}{8m L^2} $$
More Information
This formula provides the energy difference due to the spacing of energy levels in a quantum box, and it reflects how electrons can absorb or emit energy when transitioning between these quantized states. Understanding such transitions is crucial in fields like spectroscopy and quantum chemistry.
Tips
- Forgetting to square the principal quantum number (n) when calculating energy levels.
- Confusing the order of states, leading to incorrect calculations of energy transitions.
- Missing out on simplifying terms, resulting in unnecessarily complicated expressions.
