Real-Life Examples of Quantum Mechanic

Combination of de Broglie wavelength and mass–energy equivalence gone wrong?
I tried to combine the mass–energy equivalence for a particle with mass,
$E=\sqrt{(m{c}^2{)}^2+(pc{)}^2}=\sqrt{(m{c}^2{)}^2+(\mathrm{\gamma <!--\; \gamma \; -->}mvc{)}^2}$
with de Broglie wavelength,
$\mathrm{\lambda <!--\; \lambda \; -->}={\displaystyle \frac{h}{p}}={\displaystyle \frac{h}{\mathrm{\gamma <!--\; \gamma \; -->}mv}}.$
I get this equation:
$E={\displaystyle \frac{h{c}^2}{\mathrm{\lambda <!--\; \lambda \; -->}v}}.$
This does not seem right, since the equations suggest the energy increases as the speed slow down which is not the case. But I can't see what I did wrong, either. Can someone help me?

Are De Broglie relations only applicable to particles that have zero potential energy?
De Broglie relations are always written as:
$E=h\mathrm{\nu <!--\; \nu \; -->}$
$p=\frac{h}{\mathrm{\lambda <!--\; \lambda \; -->}}$
However, it doesn't make sense when you have waves that are eigenstates of a Hamiltonian operator with a non-zero potential. That is because that would give us a direct relation between momentum and energy: $\frac{E}{p}=v_p$, where $v_p$ is the phase velocity. This doesn't seem to make sense to me. So, would these equations be the same with the electron of the hydrogen atom, for example?

Wien's displacement law in frequency domain
When I tried to derive the Wien's displacement law I used Planck's law for blackbody radiation:
$I_{\mathrm{\nu <!--\; \nu \; -->}}=\frac{8\mathrm{\pi <!--\; \pi \; -->}{\mathrm{\nu <!--\; \nu \; -->}}^2}{c^3}\frac{h\mathrm{\nu <!--\; \nu \; -->}}{e^{h\mathrm{\nu <!--\; \nu \; -->}/k_{b}T}-<!--\; -\; -->1}$
Asking for maximum:
$\frac{d{I}_{\mathrm{\nu <!--\; \nu \; -->}}}{d\mathrm{\nu <!--\; \nu \; -->}}=0:0=\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\nu <!--\; \nu \; -->}}(\frac{{\mathrm{\nu <!--\; \nu \; -->}}^3}{e^{h\mathrm{\nu <!--\; \nu \; -->}/k_{b}T}-<!--\; -\; -->1})=\frac{3{\mathrm{\nu <!--\; \nu \; -->}}^2(e^{h\mathrm{\nu <!--\; \nu \; -->}/k_{b}T}-<!--\; -\; -->1)-<!--\; -\; -->{\mathrm{\nu <!--\; \nu \; -->}}^{3}h/k_{b}T\cdot <!--\; \cdot \; -->e^{h\mathrm{\nu <!--\; \nu \; -->}/k_{b}T}}{(e^{h\mathrm{\nu <!--\; \nu \; -->}/k_{b}T}-<!--\; -\; -->1{)}^2}$
It follows that numerator has to be $0$ and looking for $\mathrm{\nu <!--\; \nu \; -->}>0$
$3(e^{h\mathrm{\nu <!--\; \nu \; -->}/k_{b}T}-<!--\; -\; -->1)-<!--\; -\; -->h\mathrm{\nu <!--\; \nu \; -->}/k_{b}T\cdot <!--\; \cdot \; -->e^{h\mathrm{\nu <!--\; \nu \; -->}/k_{b}T}=0$
Solving for $\mathrm{\gamma <!--\; \gamma \; -->}=h\mathrm{\nu <!--\; \nu \; -->}/k_{b}T$:
$3(e^{\mathrm{\gamma <!--\; \gamma \; -->}}-<!--\; -\; -->1)-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}{e}^{\mathrm{\gamma <!--\; \gamma \; -->}}=0\to <!--\; \to \; -->\mathrm{\gamma <!--\; \gamma \; -->}=2.824$
Now I look at the wavelength domain:
$\mathrm{\lambda <!--\; \lambda \; -->}=c/\mathrm{\nu <!--\; \nu \; -->}:\mathrm{\lambda <!--\; \lambda \; -->}=\frac{hc}{\mathrm{\gamma <!--\; \gamma \; -->}{k}_b}\frac{1}{T}$
but from Wien's law $\mathrm{\lambda <!--\; \lambda \; -->}T=b$ I expect that $hc/\mathrm{\gamma <!--\; \gamma \; -->}{k}_b$ is equal to $b$ which is not:
$\frac{hc}{\mathrm{\gamma <!--\; \gamma \; -->}{k}_b}=0.005099$, where $b=0.002897$
Why the derivation from frequency domain does not correspond the maximum in wavelength domain?
I tried to justify it with chain rule:
$\frac{dI}{d\mathrm{\lambda <!--\; \lambda \; -->}}=\frac{dI}{d\mathrm{\nu <!--\; \nu \; -->}}\frac{d\mathrm{\nu <!--\; \nu \; -->}}{d\mathrm{\lambda <!--\; \lambda \; -->}}=\frac{c}{{\mathrm{\nu <!--\; \nu \; -->}}^2}\frac{dI}{d\mathrm{\nu <!--\; \nu \; -->}}$
where I see that $c/{\mathrm{\nu <!--\; \nu \; -->}}^2$ does not influence where $d{I}_{\mathrm{\lambda <!--\; \lambda \; -->}}/d\mathrm{\lambda <!--\; \lambda \; -->}$ is zero.

Uses of Wien's law of displacement
Wiens's displacement law says
${\lambda }_{\text{max}}T=\text{a constant}$
So if I have the ${\lambda }_{\text{max}}$, I can find the temperature of a star. But if I have the temperature, is there any point in calculating ${\lambda }_{\text{max}}$? What information does that give us of the star, besides temperature?

Deeper underlying explanation for color-shift in Wien's Law?
Suppose we have a blackbody object, maybe a star or a metal (although I understand neither of these are actually blackbody objects, to some extent my understanding is they can approximate one). According to Wien's Law, as temperature increases, peak wavelength will decrease, so the color we observe will "blueshift." Despite having reached Wien's Law in my research, my understanding is that this is in fact not an answer to why the blueshift occurs, but just a statement that the blueshift does occur.
So, is there some deeper reason why blackbodies peak wavelength emitted decreases with temperature?

De Broglie wavelength of an electron
If an electron which already possesses some kinetic energy of $X\mathrm{e}\mathrm{V}$ is further accelerated through a potential difference of $X\prime \mathrm{V}$, then is it correct to state that the electron now has a total kinetic energy of $(X+X\prime )\mathrm{e}\mathrm{V}$? Using this information, am I allowed to substitute this value in the de Broglie equation to find the wavelength of the electron?

What is $E$ in below equation? Does it represent total energy or kinetic energy(De-broglie equation for uncharged particle like neutron.)
De-broglie equation for uncharged particle:
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{\sqrt{2mE}}$
Where, $\mathrm{\lambda <!--\; \lambda \; -->}$ = wavelength
$h$ = planks constant
$m$ = mass of uncharged particles

Wien's Displacement Law and reradiation of LWIR by $\mathrm{C}{\mathrm{O}}_2$
Consider Wien's Displacement Law (if I understand correctly): λ = b / T Where, λ = Peak Wavelength b = 0.028977 mK (Wien's constant) T = Temperature
According to this law $\mathrm{C}{\mathrm{O}}_2$ molecules can only absorb and reradiate the Long Wave I.R. frequency (radiated from Earth) in their 15 micrometre (μm) spectrum at a temperature of -80 degrees Celcius. The link with temperature seems crucial to me since the only part of our athmosphere cold enough for this to happen is the Mesosphere at about 50-80 km from Earth. Troposphere and Stratosphere are not cold enough (coldest temperature is about -55 degrees Celsius). In these parts of our atmosphere it is just not cold enough for $\mathrm{C}{\mathrm{O}}_2$ to reradiate I.R. wavelengths back to Earth. Then why is $\mathrm{C}{\mathrm{O}}_2$ considered a greenhouse gas in our Tropo- or Stratosphere? Am I missing out on something here?
In the Mesosphere, however, temperature can drop to over -100 degrees Celsius. In this part of the atmosphere $\mathrm{C}{\mathrm{O}}_2$ can reradiate LWIR back to Earth. But the problem is that the air is so thin, there are hardly any molecules of $\mathrm{C}{\mathrm{O}}_2$
What am I missing here in my reasoning?

Derivation of time-dependent Schrödinger equation from De Broglie hypothesis
In a quantum mechanics script I'm reading, the Schrödinger equation is "derived" (more precisely, motivated) by the De Broglie hypothesis. It starts at
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{2\mathrm{\pi <!--\; \pi \; -->}h}{p}$
$\mathrm{\omega <!--\; \omega \; -->}=\frac{E}{h}$
then takes the partial derivatives of the wave $\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=\mathrm{\Psi <!--\; \Psi \; -->}(0,0)e^{\frac{2\mathrm{\pi <!--\; \pi \; -->}ix}{\mathrm{\lambda <!--\; \lambda \; -->}}-<!--\; -\; -->it\mathrm{\omega <!--\; \omega \; -->}}$
$\begin{array}{}\text{(1)}& \frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=\frac{ip}{h}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)\end{array}$
$\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=-<!--\; -\; -->i\frac{E}{h}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)$
With the non-relativistic free particle $E=\frac{1}{2m}{p}^2$ one gets
$\begin{array}{}\text{(2)}& ih\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=E\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=\frac{p^2}{2m}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)\end{array}$
From there, they miraculously get the time-dependent Schrödinger equation. I cannot understand this step. If I insert formula (1) for $p^2$ in (2), I get something with $(\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}\mathrm{\Psi <!--\; \Psi \; -->}{)}^2$, which is not the second partial derivative of $\mathrm{\Psi <!--\; \Psi \; -->}$ with respect to $x$.
Any hints?

Is the relation c=νλ valid only for Electromagnetic waves?
What is the validity of the relation $c=\mathrm{\nu <!--\; \nu \; -->}\mathrm{\lambda <!--\; \lambda \; -->}$? More specifically, is this equation valid only for Electromagnetic waves?
I read this statement in a book, which says:
"de Broglie waves are not electromagnetic in nature, because they do not arise out of accelerated charged particle."
This seems correct, but arises a doubt in my mind.
Suppose I find out the wavelength of a matter wave (or de Broglie wave) using de Broglie's wave equation:
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$
Now, can I use $c=\mathrm{\nu <!--\; \nu \; -->}\mathrm{\lambda <!--\; \lambda \; -->}$ to find out the frequency of the wave?

Proof of de Broglie wavelength for electron
According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is $\mathrm{\lambda <!--\; \lambda \; -->}=h/mv$
The proof of this is given in my textbook as follows:
1.De Broglie first used Einstein's famous equation relating matter and energy,
$E=m{c}^2,$
where $E=$ energy, $m=$ mass, $c=$ speed of light.
2.Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation,
$E=h\mathrm{\nu <!--\; \nu \; -->},$
where $E=$ energy, $h=$ Plank's constant ($6.62607\times <!--\; \times \; -->{10}^{-<!--\; -\; -->34}\mathrm{J}\mathrm{s}$), $\mathrm{\nu <!--\; \nu \; -->}=$ frequency.
3.Since de Broglie believes particles and wave have the same traits, the two energies would be the same:
$m{c}^2=h\mathrm{\nu <!--\; \nu \; -->}.$
$m{c}^2=h\mathrm{\nu <!--\; \nu \; -->}.$
4.Because real particles do not travel at the speed of light, de Broglie substituted $v$, velocity, for $c$, the speed of light:
$m{v}^2=h\mathrm{\nu <!--\; \nu \; -->}.$
I want a direct proof without substituting $v$ for $c$. Is it possible to prove directly $\mathrm{\lambda <!--\; \lambda \; -->}=h/mv$ without substituting $v$ for $c$ in the equation $\mathrm{\lambda <!--\; \lambda \; -->}=h/mc$?

What is the most intense wavelength and frequency in the spectrum of a black body?
Planck's Law is commonly stated in two different ways:
$u_{\mathrm{\lambda <!--\; \lambda \; -->}}(\mathrm{\lambda <!--\; \lambda \; -->},T)=\frac{2h{c}^2}{{\mathrm{\lambda <!--\; \lambda \; -->}}^5}\frac{1}{e^{\frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}kT}}-<!--\; -\; -->1}$
$u_{\mathrm{\nu <!--\; \nu \; -->}}(\mathrm{\nu <!--\; \nu \; -->},T)=\frac{2h{\mathrm{\nu <!--\; \nu \; -->}}^3}{c^2}\frac{1}{e^{\frac{h\mathrm{\nu <!--\; \nu \; -->}}{kT}}-<!--\; -\; -->1}$
We can find the maximum of those functions by differentiating those equations with respect to $\mathrm{\lambda <!--\; \lambda \; -->}$ and to $\mathrm{\nu <!--\; \nu \; -->}$, respectively. We get two ways to write Wien's Displacement Law:
${\mathrm{\lambda <!--\; \lambda \; -->}}_{\text{peak}}T=2.898\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->3}m\cdot <!--\; \cdot \; -->K$
$\frac{{\mathrm{\nu <!--\; \nu \; -->}}_{\text{peak}}}{T}=5.879\cdot <!--\; \cdot \; -->{10}^{10}Hz\cdot <!--\; \cdot \; -->K^{-<!--\; -\; -->1}$
We see that ${\mathrm{\lambda <!--\; \lambda \; -->}}_{\text{peak}}\ne <!--\; \ne \; -->\frac{c}{{\mathrm{\nu <!--\; \nu \; -->}}_{\text{peak}}}$. So what frequency or wavelength is actually detected by an optical instrument most intensely when analyzing a black body? If they are ${\mathrm{\lambda <!--\; \lambda \; -->}}_{\text{peak}}$ and ${\mathrm{\nu <!--\; \nu \; -->}}_{\text{peak}}$, how is ${\mathrm{\lambda <!--\; \lambda \; -->}}_{\text{peak}}\ne <!--\; \ne \; -->\frac{c}{{\mathrm{\nu <!--\; \nu \; -->}}_{\text{peak}}}$?

Why $c$ needs to be added in numerator and denominator while solving a de Broglie equation word problem?
Calculate the de Broglie wavleength of a 0.05 eV ("theremal") neutron.Making a nonrelativistic calculation,
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}=\frac{h}{\sqrt{2m_{0}K}}=\frac{hc}{\sqrt{2\left(m_0{c}^2\right)K}}=\frac{12.4\times <!--\; \times \; -->{10}^3\mathrm{e}\mathrm{V}\cdot <!--\; \cdot \; -->\text{Å<!-- Å -->}}{\sqrt{2(940\times <!--\; \times \; -->{10}^6\mathrm{e}\mathrm{V})\left(0.05\mathrm{e}\mathrm{V}\right)}}=1.28\text{Å<!-- Å -->}$
I am confused as to why it is necessary to add $c$ in the numerator and denominator (red equation) while solving a word problem like this?

What is the difference between Wien's Displacement Law for peak frequency vs peak wavelength?
While doing research for a high school report I came across the fact that WDL actually has two forms, one for peak frequency and one for peak wavelength, and that these two forms are not the same and can not be used interchangeably.
My question is why peak frequency isn't the same as peak wavelength? That is, since wavelength is directly determined by frequency (since frequency = speed of light divided by wavelength), there is a one-to-one correspondence between a given wavelength and its frequency. Therefore why doesn't a peak in frequency correspond to a peak in wavelength, and visa versa (meaning that the two forms of WDL could be used interchangeably)?
I know that this question was already posted elsewhere, but I did not understand the answers. Since I am a high school student, complicated terminology can fly right over my head, so I would greatly appreciate it if someone could take the time to explain it clearly and simply (i.e. no monster equations).

Blackbody radiation, de Broglie equation, and lightwaves to be shifted left
I'm having a hard time figuring this out.
1.Say we heated a lead ball to 1,000 Kelvin. Not all of the particles are at the exact same temperature--some parts are a little hotter, some are a little cooler. But for now, let’s assume that it follows a normal distribution that is centered at 1,000K.
2.The heat (or movement of the molecules) causes it to emit light.
3.Because the light photons have to be discrete (you can't have a 1/2 photon), this causes the observed light wavelengths to be shifted left.
4.This means we observe more red light than we might otherwise expect.
5.This is a long wind up to my specific question--does a particle vibrating at a specific frequency emit light at the same frequency? (i.e. a particle vibrating at 4.3 MhZ emits light at 4.3 Mhz). Because it seems like the whole thing hinges on that.
I mention this because I asked a physics teacher this, and he said, “No, the particles emit light following the de Broglie equation.” This would mean that the light emitted ignores the frequency and instead is based solely on its momentum. But, if this were true, then I would assume it would emit light in a standard distribution of frequencies as opposed to the left-skewed distribution that is actually observed.
Any help on this would be greatly appreciated!

De-Broglie equation for other particles except photon
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?

Wavelength and relativity
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 ?

Special Relativity, Louis de Broglie Equation Dilemma
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 $E=mc²$ to form $\frac{hc}{\lambda }=mc²$ 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 $E=mc²$ because the $E=mc²$ is for particles while the former is for waves.
Is my understanding correct$?$ Please help and thanks in advance$!$

Which formula for the de Broglie wavelength of an electron is correct?
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: $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$, 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: $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{hc}{E}$. 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 $400MeV$

Wien's fifth power Law and Stephan Boltzmann's fourth power laws of emissive power
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?