How do you prove ##\lambda = \frac {2L} {n} ## given only L?

In summary, to prove that \(\lambda = \frac{2L}{n}\) given only \(L\), one can start by recognizing that \(\lambda\) represents the wavelength of a wave, while \(L\) is typically the length of a vibrating string or medium. The relationship arises from the fundamental frequency of standing waves, where \(n\) denotes the mode number or harmonic. For a string fixed at both ends, the wavelength is inversely proportional to the number of nodes, leading to the formula \(\lambda = \frac{2L}{n}\), which can be derived from the conditions of standing wave patterns along the length \(L\).
  • #1
Danielk010
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Homework Statement
A free proton moves back and forth between rigid walls separated by a distance L.

If the proton is represented by a one-dimensional standing de Broglie wave with a node at each wall, show that the allowed values of the de Broglie wavelength are given by the given equation where n is a positive integer.
Relevant Equations
##\lambda = \frac {2L} {n} ##
##\lambda = \sqrt { \frac {(hc)^2} {2mc^2k} } ##, where k is the kinetic energy
## (hc)^2 = 1240 (ev *nm)^2 ##
Since I know from the equation the type of particle and the distance L, I thought of equating the first relevant equation to the second equation. Since n = 1, 2, 3 ..., I thought by equating the two equations I could get k = 1, 4, 9... and have the two constants equal each other. The two constants did not equal each other, so I am a bit stuck on where to go from here or where to start. I got an equation for kinetic energy in terms of n from my previous attempt, but I don't know the quantum number, n, nor the kinetic energy, k. Thank you for any help that can provided.
 
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  • #2
Danielk010 said:
I got an equation for kinetic energy in terms of n from my previous attempt, but I don't know the quantum number, n, nor the kinetic energy, k.
What previous attempt? Did you solve the Schrodinger equation for a particle in a box? You don't need to know ##n## because it is part of ##K##. This is a "show that" kind of problem. You don't have to find any numbers.
 
  • #3
kuruman said:
What previous attempt? Did you solve the Schrodinger equation for a particle in a box? You don't need to know ##n## because it is part of ##K##. This is a "show that" kind of problem. You don't have to find any numbers.
My previous attempt was trying to equate de Broglie's equation from the equation given in the question. I did not solve Schrodinger equation for a particle in a box. My class went over Schrodinger's equation involving a infinite barrier quantum well, I am not sure if that is what you are referring to. Looking back on my notes, I think I figured it out. Thank you for the help.
 
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  • #4
Danielk010 said:
My previous attempt was trying to equate de Broglie's equation from the equation given in the question. I did not solve Schrodinger equation for a particle in a box. My class went over Schrodinger's equation involving a infinite barrier quantum well, I am not sure if that is what you are referring to. Looking back on my notes, I think I figured it out. Thank you for the help.
Yes, I was referring to an infinite barrier well a.k.a. particle in a box. I'm glad you figured it out by yourself.
 
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FAQ: How do you prove ##\lambda = \frac {2L} {n} ## given only L?

How do you prove ##\lambda = \frac {2L} {n}## given only L?

To prove ##\lambda = \frac {2L} {n}## given only the length L, you need to understand the context of the problem. This equation typically arises in the study of standing waves on a string or in a pipe. The derivation involves understanding the relationship between the wavelength (##\lambda##), the length of the medium (L), and the harmonic number (n). However, to prove this, you usually need additional information such as boundary conditions or the harmonic series relationship.

What is the significance of the variables in the equation ##\lambda = \frac {2L} {n}##?

In the equation ##\lambda = \frac {2L} {n}##, ##\lambda## represents the wavelength of the wave, L is the length of the medium (such as a string or pipe), and n is the harmonic number, which is an integer representing the mode of vibration. The equation shows that the wavelength is inversely proportional to the harmonic number and directly proportional to twice the length of the medium.

Why is the factor of 2 present in the equation ##\lambda = \frac {2L} {n}##?

The factor of 2 is present because it accounts for the fact that a standing wave in its nth harmonic forms n/2 wavelengths over the length L. For instance, in the first harmonic (n=1), there is half a wavelength in the length L, so ##\lambda = 2L##. In the second harmonic (n=2), there is one full wavelength in the length L, so ##\lambda = L##, and so on.

Can you derive ##\lambda = \frac {2L} {n}## step by step?

Yes, here's a simple derivation: For a string fixed at both ends, the fundamental frequency (first harmonic) has a wavelength ##\lambda_1 = 2L##. For the second harmonic, the string has one full wavelength fitting within the length L, so ##\lambda_2 = L##. Generally, for the nth harmonic, the string fits n/2 wavelengths within the length L. Therefore, ##n(\lambda/2) = L##, which rearranges to ##\lambda = \frac{2L}{n}##.

What are the practical applications of the equation ##\lambda = \frac {2L} {n}##?

This equation is widely used in physics and engineering, particularly in acoustics and the design of musical instruments. It helps in determining the frequencies of the harmonics produced by a

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