Recent questions in Quantum Mechanics

de Broglie Equation
Answered

Fescoisyncsibgyp8b
2022-05-13

If E=hf is applicable for electron and other particles, the De Broglie wavelength should be λ=hv/pc. Because, mc^2=hf which implies mc^2=hc/λ which implies m=h/λc and thus λ=hv/pc. But I have found in my text book that λ=h/p is applicable not only for photon but also for all particle. But how can λ=h/p=h/mv be applicable for all particle?

de Broglie Equation
Answered

vilitatelp014
2022-05-13

From de Broglie equation λ=h/p. But p=mv and velocity is a relativistic quantity so also wavelength is relative ? In other words does wavelength depends on the reference frame ?

de Broglie Equation
Answered

studovnaem4z6
2022-05-13

While learning atomic structure I stumbled upon a very unusual doubt.

As we know that the energy of a wave is given by the equation: $E=\frac{hc}{\lambda}$ and Louis de broglie wave equation is given by the equation ${\lambda}_{B}=\frac{h}{p}$. My doubt is that, that is ${\lambda}_{B}=\lambda $. Do the ${\lambda}_{B},\lambda $ represent the same thing $?$

My teacher equated $E=\frac{hc}{\lambda}$ and $E=mc\xb2$ to form $\frac{hc}{\lambda}=mc\xb2$ and rearranged to form $\lambda =\frac{h}{mc}$ and then replaced $\lambda $ by ${\lambda}_{{\rm B}}$ and $c$ by $v$ for general formula and derived the Louis de broglie equation. This created my doubt in first place and I created another doubt that whether the equation for energy of wave is valid for relativistic equation of $E=mc\xb2$ because the $E=mc\xb2$ is for particles while the former is for waves.

Is my understanding correct$?$ Please help and thanks in advance$!$

de Broglie Equation
Answered

Jaiden Bowman
2022-05-10

So, I have my exams in physics in a week, and upon reviewing I was confused by the explanation of de Broglie wavelength of electrons in my book. Firstly, they stated that the equation was: $\lambda =\frac{h}{p}$, where $h$ is the Planck constant, and $p$ is the momentum of the particle. Later, however, when talking about electron diffraction and finding the angles of the minima, the author gave the formula equivalent to that for light: $\lambda =\frac{hc}{E}$. Now, what I don't understand is if it is simply a mistake made by the author, or whether a different formula have to be used for electron diffraction, as the two formulae are very clearly not equivalent. In the latter case, I don't understand why the formula would be different. I greatly appreciate the help, as the exams are really close, and I would like to make sure I get this right!

Edit: I was told that pictures of text are taking away from the readability of the posts, and thus they were removed. Essentially, the difference between the two cases are that in the first case, the proton did not have any significantly large kinetic energy, while in the second example, the kinetic energy was $400\text{}MeV$

Wien's displacement law
Answered

Adelyn Rodriguez
2022-05-10

Wien's fifth power law says that emissive power is proportional to the temperature raised to the fifth power. On the other hand, the Stefan–Boltzmann law says emissive power is proportional to the temperature raised to the fourth power. How can both of these be true?

Wien's displacement law
Answered

hovudverkocym6
2022-05-10

It is known that for perfect blackbodies,

$\lambda T=c$

where $\lambda =\text{peak wavelength}$

$T=\text{Absolute temperature}$

$c=\text{Wien's constant}$

But this is for perfect blackbodies only, which have no theoretical existence. Does a similar formula exist for real bodies, which expresses $\lambda T$ in terms of its emissitivity $\u03f5$? I googled it, but found no relevant results.

Wien's displacement law
Answered

Deshawn Cabrera
2022-05-09

Why does the area under Wien's displacement graph give Stefan-Boltzmann law for a black body?

I couldn't find any proof of this. (I could just find this expression). I am not aware of the function of Wien's displacement graph as well (I just know that it is between Intensity and wavelength emitted by a black body).

Is there a mathematical way to prove this?

Wien's displacement law
Answered

bedblogi38am
2022-05-09

This is a question relating to Wien's displacement law for the Planck function. As we all know frequency and wavelength are related to the speed of light by:

$\nu \lambda =c$

However, why is it that:

${\nu}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}{\lambda}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}\ne c$

Any explanations would be very much appreciated.

To all of the people wanting to know where this statement came from. It hasn't come from anywhere specific, is it a well known fact of the Planck function. ${\lambda}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}=0.290{T}^{-1}$ cm K and ${\nu}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}=5.88\times {10}^{10}T$ Hz K${}^{-1}$

Wien's displacement law
Answered

arbixerwoxottdrp1l
2022-05-09

Wien's Displacement Law stated that for a blackbody emitting radiation,

${\lambda}_{max}={\displaystyle \frac{1}{T}}$

where $T$ is the temperature of the body and ${\lambda}_{max}$ is the maximum wavelength of radiation emitted.

Due to the relationship between wavelength, frequency and the speed of light, a value of maximum wavelength would give a value of minimum frequency, and vice versa.

I then saw on the Wikipedia page for Wien's Displacement Law that

${f}_{max}={\displaystyle \frac{\alpha {k}_{B}T}{h}},$

where $\alpha =\mathrm{2.82...}$, ${k}_{B}$ is Boltzmann's Constant, $T$ is the temperature of the body and $h$ is Planck's Constant.

How can this relationship for maximum frequency be shown?

de Broglie Equation
Answered

Hailee Stout
2022-05-09

Do the properties of the waves (wavelength,frequency) emitted by a particle or object depend upon the velocity, or as to say its kinetic energy? Is the De Broglie equation $E=h\nu $ applicable to matter waves as well?

de Broglie Equation
Answered

rynosluv101wopds
2022-05-08

Just wondering if anyone can help me understand the basic principle of quantum theory.

De Broglie's equation allows one calculate the wave length of the physical object, following the fundamental wave-particle duality of quantum theory.

$\lambda =h/mv$

Since velocity $v$ is always relative to the reference frame of observer, does it imply that the wave property is not inherent but displays itself differently to different observers?

Wien's displacement law
Answered

Brooklynn Hubbard
2022-05-08

My attempt to solve this question was to use the following assumptions:

1.Temperature of CMB photon today is 2.725 K (will use value of 3 K here)

2.Temperature of CMB photon when it was first emitted is 3000 K

3.A factor of x1000 in temperature decrease results in a factor of x1000 in wavelength increase. (According to Wien's displacement law)

Does this mean that the source of the CMB photon that just reached me today, was actually 13.7 billion light years / 1000 = 13.7 million light years away from me when it first emitted the photon?

de Broglie Equation
Answered

Kiersten Hodge
2022-05-08

Louis de Broglie suggested that for microparticles like electrons, wave-like properties can be applied in order to explain some phenomena. Schrodinger wrote down an equation, a wave equation, describing these waves. What I do not understand is why is Schrodinger's contribution so important; if the concept of wave-like property of an electron was known already, then why writing a mere wave equation was an important step?

de Broglie Equation
Answered

kwisangqaquqw3
2022-05-08

I've just started learning about the double slit experiment (just in the short appendix section in Schroeder's Thermal Physics), and I'm extremely confused by this one thing:

In it, out of basically nowhere he pulls out the De Broglie equation, that λ = h/p.

I've studied double slit diffraction before, and I've been trying to connect them in order to understand what this wavelength actually means.

In double slit diffraction, when the wavelength is larger, the diffraction "stripes" that form on the wall appear further apart. They also appear larger.

y = $\frac{m\lambda L}{d}$ (approx, considering the distance to the screen is really large and thus almost parallel rays (drawn out waves) can have a path difference and interfere)

If we were to make the wavelength extremely small, that would mean that anything a little off-center would interfere, so the smaller the wavelength, the closer together the "stripes" on the wall would be.

Now, when we connect the 2 equations, this means that the faster the electrons are moving (the smaller their wavelength) the more places they will interfere on the wall, and therefore there will be a lesser distance between adjacent places where the electrons hit (bright spots) and places where they don't (dark spots).

The way I'm interpreting this is that the smaller the "wavelength" of the electron, the more the probability it has to have been in different places at the same time, that is, the less we can know its position. That's why more stripes will appear on the detecting screen because there are more positions which the electron could've been in, and since its technically in all of them at the same time while it travels, it can interfere with itself more.

Is this interpretation correct? Does a faster momentum (a smaller wavelength) mean that the electron literally is at more places at the same time while it travels from the electron gun, through the slits, and to the wall? Thank you!

de Broglie Equation
Answered

Eve Dunn
2022-05-08

I was curious about the famous $p=\hslash k$ equation. In high school I think you are just exposed to this equation with the explanation of "something something matter waves." But early in a undergraduate QM course you solve the time-independent Schrodinger's equation for a free particle in 1D and get the following solution:

$\frac{-{\hslash}^{2}}{2m}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{x}^{2}}\psi =E\psi \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\psi ={e}^{\pm ikx}$ and $E=\frac{{\hslash}^{2}{k}^{2}}{2m}$

where I believe $k$ is just defined by the $E(k)$ equation. And then do we just realize that the $E(k)$ is the classical ${p}^{2}/2m$ equation if we set $p=\hslash k$, and thus we have "derived" $p=\hslash k$ or do we "know" $p=\hslash k$ beforehand and this just confirms it?

de Broglie Equation
Answered

Stoyanovahvsbh
2022-05-08

I'm trying to understand how to write Heisenberg uncertainty principle in 3 dimensions. What I mean by that is to prove something of the form $f(\mathrm{\Delta}{p}_{x},\mathrm{\Delta}{p}_{y},\mathrm{\Delta}{p}_{z},\mathrm{\Delta}x,\mathrm{\Delta}y,\mathrm{\Delta}z)\ge A$

This is what I got: The unknown volume that a single particle can be in is $\mathrm{\Delta}V=\mathrm{\Delta}x\mathrm{\Delta}y\mathrm{\Delta}z$. The uncertainty in the size of the momentum is $\mathrm{\Delta}p=\sqrt{\mathrm{\Delta}{p}_{x}^{2}+\mathrm{\Delta}{p}_{y}^{2}+\mathrm{\Delta}{p}_{z}^{2}}$

Now this is where I get stuck. In my textbook, for the 1d case, they used De-Broglie equation for connecting the uncertainty of the particle wavelength and its momentum along the $x$-axis. But does De-Broglie equation is correct per axis or for the size of the vectors?

Thanks for you help

Wien's displacement law
Answered

Thaddeus Sanders
2022-05-08

According to Wikipedia:

"Long-wavelength infrared (8–15 µm, 20–37 THz, 83–155 meV): The "thermal imaging" region, in which sensors can obtain a completely passive image of objects only slightly higher in temperature than room temperature - for example, the human body - based on thermal emissions only and requiring no illumination such as the sun, moon, or infrared illuminator. This region is also called the "thermal infrared"."

However, using $\frac{hc}{\lambda}={k}_{\mathrm{B}}T$, the temperature range 288–308 K (15–35 °C) is equivalent to 50–46.7 µm, while 8–15 µm is equivalent to 1800–960 K (using the same equation).

de Broglie Equation
Answered

Annabel Sullivan
2022-05-08

I was wondering if using the De Broglie equation

$\lambda =\frac{h}{p}$

for object traveling at really high speeds would result in a significant error. For example if an object travelled at $0.02c$ would the error be negligible? How can I calculate the uncertainty in the result?

Wien's displacement law
Answered

studovnaem4z6
2022-05-07

I know this might be a silly question, but is it necessary to know Planck's Law in order to show that ${\lambda}_{max}\propto \frac{1}{T}$? If you set

$u(\lambda ,T)=\frac{f(\lambda T)}{{\lambda}^{5}}$

then

$\frac{\mathrm{\partial}u}{\mathrm{\partial}\lambda}=\frac{1}{{\lambda}^{5}}\frac{\mathrm{\partial}f}{\mathrm{\partial}\lambda}-\frac{5}{{\lambda}^{4}}f=0$

$\frac{\frac{\mathrm{\partial}f}{\mathrm{\partial}\lambda}}{f}=5\lambda $

But I am stuck here because if I integrate the L.H.S.

$\int \frac{\frac{\mathrm{\partial}f}{\mathrm{\partial}\lambda}}{f}d\lambda =\mathrm{log}(f(\lambda T))=\frac{5}{2}{\lambda}^{2}$

de Broglie Equation
Answered

Waylon Mcbride
2022-05-07

$p=mv$

where $m$ is the mass of an electron, and $v$ is its velocity. In this case, since $v$ is a vector, it's clear that the momentum will be also a vector.

However if the momentum is a vector quantity (and it is), what is the direction of the electron's momentum given by the de Broglie relation

$p=h/\lambda \phantom{\rule{0ex}{0ex}}p=\hslash k$

if the Planck constant $h$ is scalar and the wavelength $\lambda $ is also scalar. Similarly the reduced Planck constant $\hslash $ is scalar and the wavenumber $k=2\pi /\lambda $ is also scalar.

Quantum mechanics problems and solutions belong to one of the most complex aspects of Physics because it takes several disciplines to get things done. It is exactly why we have a plethora of different questions that are asked by engineers, designers, programmers, data scientists, and students who do not have sufficient skills in Physics. Take your time to study quantum mechanics examples and explore the answers that have been offered. There are also de Broglie equation problems that focus on the wave properties of matter and the nature of the electrons that will help you. If you need to experiment with the other types of Quantum Mechanics and explore the radiation physics, make sure to che