show that the de Broglie wavelenght of a particle in one dimensional box in the first excited state is equal to the lenght of the box
Answers
Answer:
The allowed wavelengths of the standing waves of a giant "classical" spring are found to be "quantized."
Ingredients: giant spring
Procedure: A complete recipe follows.
1. Attach one end of the spring to a fixed point.
2. Extend the spring to straighten it, developing substantial tension.
3. Add vibrational energy to the spring by moving the opposite end.
4. Observe the allowed wavelengths of the standing waves that are created in the spring.
Understanding: The oscillation of the giant spring reminds us of our many experiences with waves formed on vibrating strings, including jump ropes and the strings of a guitar. Quickly moving one end of the spring up and down sets a wave pulse moving along the length of the spring. The propagating wave reflects off the far end of the spring, returning to our hands, reflecting again, and so on, until it dissipates. We find that we can create waves of practically any length, by varying how quickly we "pulse" the spring.
If we add energy to the spring, but keeping the ends fixed, we create standing waves. For the lowest energy wave, the two ends of the spring are held fixed and the amplitude of the wave has one maximum that oscillates up and down with a fixed frequency. The wavelength of the wave is twice the length of the spring. Adding energy to the spring induces more oscillations in the wave. The points along the wave where the amplitude is zero are called nodes. A wave with one node in the center has a wavelength equal to the length of the spring. A wave with three nodes has a wavelength equal to one-half the length of the spring. The ends of the spring don't count as nodes.
It is not hard to determine all of the possible wavelengths for the standing waves of a spring of length L. The longest wavelength for a standing wave is λ=2L. Anything longer and the wave will not have fixed points at the beginning and end of the spring. We can also have a wavelength of λ=L, where one full oscillation of the wave fits on the spring. Or we can have a wavelength of λ=2L/3, where 3 half-wavelengths fit along the spring. And so on. In general, there are an infinite number of allowed wavelengths
λn = 2L/n n = 1,2,3...
We can call n/2 the wavenumber because it is the number of wavelengths λn that fit in a box of length L. But what does that have to do with the discrete nature of the emission and absorption line spectra of the hydrogen atom?
A general unhappiness with the Bohr model and its orbits
Many people were unhappy with the basic physical assumptions that led to the "quantized" spectra predicted by Bohr's model of the hydrogen atom. In the Bohr model, the electron is imagined to be a point particle in a circular orbit around the proton nucleus, allowed to assume only certain values of the angular momentum
mvr = nh/2π
Quantum mechanical theory of a "particle in a box"
Explanation: