Real-Life Examples of Quantum Mechanic

Wien's Displacement Law for real bodies
It is known that for perfect blackbodies,
$\mathrm{\lambda <!--\; \lambda \; -->}T=c$
where $\mathrm{\lambda <!--\; \lambda \; -->}=\text{peak wavelength}$
$T=\text{Absolute temperature}$
$c=\text{Wien's constant}$
But this is for perfect blackbodies only, which have no theoretical existence. Does a similar formula exist for real bodies, which expresses $\mathrm{\lambda <!--\; \lambda \; -->}T$ in terms of its emissitivity $\mathrm{ϵ<!--\; ϵ\; -->}$? I googled it, but found no relevant results.

Area under Wien's displacement graph
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?

Frequency and Wavelength peak for Wien's displaement law of a blackbody
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:
$\mathrm{\nu <!--\; \nu \; -->}\mathrm{\lambda <!--\; \lambda \; -->}=c$
However, why is it that:
${\mathrm{\nu <!--\; \nu \; -->}}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}{\mathrm{\lambda <!--\; \lambda \; -->}}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}\ne <!--\; \ne \; -->c$
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. ${\mathrm{\lambda <!--\; \lambda \; -->}}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}=0.290T^{-<!--\; -\; -->1}$ cm K and ${\mathrm{\nu <!--\; \nu \; -->}}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}=5.88\times <!--\; \times \; -->{10}^{10}T$ Hz K${}^{-<!--\; -\; -->1}$

How can Wien's Displacement Law be 'changed' to a version for frequency?
Wien's Displacement Law stated that for a blackbody emitting radiation,
${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}={\displaystyle \frac{1}{T}}$
where $T$ is the temperature of the body and ${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}$ 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
$f_{max}={\displaystyle \frac{\mathrm{\alpha <!--\; \alpha \; -->}{k}_{B}T}{h}},$
where $\mathrm{\alpha <!--\; \alpha \; -->}=\mathrm{2.82...}$, $k_B$ 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?

Does the wavelenth of matter waves depend upon the kinetic energy of the particle or object?
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 $E=h\mathrm{\nu <!--\; \nu \; -->}$ applicable to matter waves as well?

How to interpret the velocity in de Broglie's equation?
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.
$\mathrm{\lambda <!--\; \lambda \; -->}=h/mv$
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?

If a CMB photon traveled for 13.7 billion years (- 374,000 years) to reach me. How far away was the source of that CMB photon when it first emitted it?
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?

Importance of Schrodinger equation
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 Wavelength interpretation
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 = ${\displaystyle \frac{m\mathrm{\lambda <!--\; \lambda \; -->}L}{d}}$ (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!

Origin of the de Broglie Equation
I was curious about the famous $p=\mathrm{\hslash <!--\; \hslash \; -->}k$ 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:
$\frac{-<!--\; -\; -->{\mathrm{\hslash <!--\; \hslash \; -->}}^2}{2m}\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}\mathrm{\psi <!--\; \psi \; -->}=E\mathrm{\psi <!--\; \psi \; -->}⟹<!--\; ⟹\; -->\mathrm{\psi <!--\; \psi \; -->}=e^{\pm <!--\; \pm \; -->ikx}$ and $E=\frac{{\mathrm{\hslash <!--\; \hslash \; -->}}^2{k}^2}{2m}$
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 $p=\mathrm{\hslash <!--\; \hslash \; -->}k$, and thus we have "derived" $p=\mathrm{\hslash <!--\; \hslash \; -->}k$ or do we "know" $p=\mathrm{\hslash <!--\; \hslash \; -->}k$ beforehand and this just confirms it?

Uncertainty Principle in 3 dimensions
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 $f(\mathrm{\Delta <!--\; \Delta \; -->}{p}_x,\mathrm{\Delta <!--\; \Delta \; -->}{p}_y,\mathrm{\Delta <!--\; \Delta \; -->}{p}_z,\mathrm{\Delta <!--\; \Delta \; -->}x,\mathrm{\Delta <!--\; \Delta \; -->}y,\mathrm{\Delta <!--\; \Delta \; -->}z)\ge <!--\; \ge \; -->A$
This is what I got: The unknown volume that a single particle can be in is $\mathrm{\Delta <!--\; \Delta \; -->}V=\mathrm{\Delta <!--\; \Delta \; -->}x\mathrm{\Delta <!--\; \Delta \; -->}y\mathrm{\Delta <!--\; \Delta \; -->}z$. The uncertainty in the size of the momentum is $\mathrm{\Delta <!--\; \Delta \; -->}p=\sqrt{\mathrm{\Delta <!--\; \Delta \; -->}{p}_x^2+\mathrm{\Delta <!--\; \Delta \; -->}{p}_y^2+\mathrm{\Delta <!--\; \Delta \; -->}{p}_z^2}$
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

Why 8–15 µm is considered "thermal infrared" if typical room temperature kT is 48 µm?
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 $\frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}=k_{\mathrm{B}}T$, 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).

Is there a significant error in using De Broglie's equation for an electron at really high speed?
I was wondering if using the De Broglie equation
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$
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?

Derivation of Wien's displacement law
I know this might be a silly question, but is it necessary to know Planck's Law in order to show that ${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}\propto <!--\; \propto \; -->\frac{1}{T}$? If you set
$u(\mathrm{\lambda <!--\; \lambda \; -->},T)=\frac{f(\mathrm{\lambda <!--\; \lambda \; -->}T)}{{\mathrm{\lambda <!--\; \lambda \; -->}}^5}$
then
$\frac{\mathrm{\partial <!--\; \partial \; -->}u}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\lambda <!--\; \lambda \; -->}}=\frac{1}{{\mathrm{\lambda <!--\; \lambda \; -->}}^5}\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\lambda <!--\; \lambda \; -->}}-<!--\; -\; -->\frac{5}{{\mathrm{\lambda <!--\; \lambda \; -->}}^4}f=0$
$\frac{\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\lambda <!--\; \lambda \; -->}}}{f}=5\mathrm{\lambda <!--\; \lambda \; -->}$
But I am stuck here because if I integrate the L.H.S.
$\int <!--\; \int \; -->\frac{\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\lambda <!--\; \lambda \; -->}}}{f}d\mathrm{\lambda <!--\; \lambda \; -->}=\log <!--\; \; -->(f(\mathrm{\lambda <!--\; \lambda \; -->}T))=\frac{5}{2}{\mathrm{\lambda <!--\; \lambda \; -->}}^2$

Direction of momentum given by the de Broglie relation
$p=mv$
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
$$\begin{gathered} p=h/\mathrm{\lambda <!--\; \lambda \; -->} \\ p=\mathrm{\hslash <!--\; \hslash \; -->}k \end{gathered}$$
if the Planck constant $h$ is scalar and the wavelength $\mathrm{\lambda <!--\; \lambda \; -->}$ is also scalar. Similarly the reduced Planck constant $\mathrm{\hslash <!--\; \hslash \; -->}$ is scalar and the wavenumber $k=2\mathrm{\pi <!--\; \pi \; -->}/\mathrm{\lambda <!--\; \lambda \; -->}$ is also scalar.

Deriving photon energy equation using $E=m{c}^2$ and de Broglie wavelength
So I was learning de Broglie wavelength today in my physics class, and I started playing around with it. I wondered if it was possible to calculate the energy of a light wave given its wavelength and speed. After rearranging a bit, I plugged it into $E=m{c}^2$, and realized I had found the equation for the energy of a photon that I learned at the beginning of my quantum mechanics unit, $E=hf$
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$
$p=\frac{h}{\mathrm{\lambda <!--\; \lambda \; -->}}$
$E=m{c}^2$
$E=\frac{p}{c}\cdot <!--\; \cdot \; -->c^2$
$E=\frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}=hf$
I have a few questions about this. Firstly, I do not understand how it can make sense to do $\frac{p}{c}$ in this context, because, as I understand it, light has no mass. How can I come to $E=hf$ using the mass of a massless object?
Secondly, as I was doing those steps, I thought I would be calculating the energy of the entire light ray. I now realize I was finding the energy of a single photon. In hindsight, this makes sense, because the energy of the entire light ray must depend on some length value, correct?
This led me to two other questions: do light rays have some finite length? how do I calculate the energy of a light ray, not just a single photon?

Deriving Wien's displacement law (Zettili)
According to Zettili, we can derive Wien's displacement law from Planck's energy density
$\stackrel{~<!--\; ~\; -->}{u}(\mathrm{\lambda <!--\; \lambda \; -->},T)=\frac{8\mathrm{\pi <!--\; \pi \; -->}hc}{{\mathrm{\lambda <!--\; \lambda \; -->}}^5}\frac{1}{e^{hc/kT\mathrm{\lambda <!--\; \lambda \; -->}}-<!--\; -\; -->1}$
where $\mathrm{\lambda <!--\; \lambda \; -->}$ is the wavelength and $T$ is the temperature. Taking the derivative of this, setting it equal to zero, and rearranging terms, one gets
$\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{\mathrm{\lambda <!--\; \lambda \; -->}}=5(1-<!--\; -\; -->e^{-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}/\mathrm{\lambda <!--\; \lambda \; -->}})$
where $\mathrm{\alpha <!--\; \alpha \; -->}=hc/(kT)$. To solve this, Zettili goes on to say that we can write
$\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{\mathrm{\lambda <!--\; \lambda \; -->}}=5-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$
so that the previously mentioned equation becomes
$5-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}=5-<!--\; -\; -->5e^{-<!--\; -\; -->5+\mathrm{ϵ<!--\; ϵ\; -->}}$
Apparently, $\mathrm{ϵ<!--\; ϵ\; -->}\approx <!--\; \approx \; -->5e^{-<!--\; -\; -->5}$ is a good answer. My first question is why do we use $5-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$ to rewrite the equation? What method is this? I've never seen that before and other resources just say that they got the values by solving numerically. My second question is why $\mathrm{ϵ<!--\; ϵ\; -->}\approx <!--\; \approx \; -->5e^{-<!--\; -\; -->5}$ when there is an extra factor of $e^{\mathrm{ϵ<!--\; ϵ\; -->}}$ in the equation? Why can we ignore that factor?

De broglie equation
What is the de Broglie wavelength? Also, does the $\mathrm{\lambda <!--\; \lambda \; -->}$ sign in the de Broglie equation stand for the normal wavelength or the de Broglie wavelength? If $\mathrm{\lambda <!--\; \lambda \; -->}$ is the normal wavelength of a photon or particle, is $\mathrm{\lambda <!--\; \lambda \; -->}\propto <!--\; \propto \; -->\frac{1}{m}$ true?
Can a wave to decrease its amplitude or energy with the increase of mass?
(My chemistry teacher told me that all matter moves in the structure of a wave and because of de Broglie's equation matter with less mass shows high amplitude and wavelength while matter with great mass shows less amplitude and wavelength. So as a result of that we see that objects like rubber balls, which have a great mass relative to the electron, move in a straight line because of its mass. However, I haven't found any relationship between wavelength and amplitude.)

If I stood next to a piece of metal heated to a million degrees, but in a perfect vacuum, would I feel hot?
A friend of mine told me that if you were to stand beside plate of metal that is millions of degrees hot, inside a 100% vacuum, you would not feel its heat. Is this true? I understand the reasoning that there is no air, thus no convection, and unless you're touching it, there's no conduction either. I'm more so asking about thermal radiation emitted by it.

Energy of photon given temperautre
I have a question asking to find the energy of a photon emitted from mater of temperature $T$. If the question had asked frequency, it would have clearly been that $E_{photon}=h\cdot <!--\; \cdot \; -->f$. I do know that classically, the energy of an oscillator is $E=K_b\cdot <!--\; \cdot \; -->T$. Is it okay to use this formula for this question?