Recent questions in Quantum Mechanics

de Broglie Equation
Answered

Sappeycuii
2022-05-18

Lets say we have an electron with known De Broglie wavelength $\lambda $. Can anyone justify or explain why we calculate its energy $E$ using 1st the De Broglie relation $\lambda =h/p$ to get momentum $p$ and 2nd using the invariant interval to calculate $E$

$\begin{array}{rl}{p}^{2}{c}^{2}& ={E}^{2}-{{E}_{0}}^{2}\\ E& =\sqrt{{p}^{2}{c}^{2}+{{E}_{0}}^{2}}\end{array}$

Why we are not alowed to do it like we do it for a photon:

$\begin{array}{r}E=h\nu =h\frac{c}{\lambda}\end{array}$

These equations return different results.

Wien's displacement law
Answered

Marissa Singh
2022-05-18

I'm studying Quantum mechanics by Bransden and Joachain and in the introduction chapter it says:

Wien showed that the spectral distribution function had to be on the form

$\rho (\lambda ,T)={\lambda}^{-5}f(\lambda T)$

where $f(\lambda T)$ is a function of the single variable $\lambda T$. It is a simple matter to show that Wien's law includes Stefan-Boltzmann law $R(T)=\sigma {T}^{4}$

One of the exercises is to show this and I cannot understand how to.

This is what I've tried:

The relationship between spectral emittance and spectral distribution is

$\rho (\lambda ,T)=\frac{4}{c}R(\lambda ,T),$

where c is the speed of light, which inserted in the above equation gives

$R(\lambda ,T)=\frac{c}{4}{\lambda}^{-5}f(\lambda T).$

Now, the total spectral emittance is the integral of $R$ over all wavelengths so

$R(T)=\frac{c}{4}\underset{0}{\overset{\mathrm{\infty}}{\int}}{\lambda}^{-5}f(\lambda T)d\lambda $

This is where I'm stuck. Can anyone help me figure this out?

de Broglie Equation
Answered

Dennis Montoya
2022-05-18

When I am deriving de Broglie wavelength for a relativistic particle using ${E}^{2}={m}^{2}{c}^{4}+{p}^{2}{c}^{2}$ and equating with $E=\frac{hc}{\lambda}$, and then putting $p=kmv$, $k$ being relativistic factor, I am getting $\lambda =\frac{h}{kmc}$ instead of $\frac{h}{kmv}$.

Is there any mistake that i am doing with equating those 2 energy equations ? Or something else ?

de Broglie Equation
Answered

dresu9dnjn
2022-05-18

I tried to combine the mass–energy equivalence for a particle with mass,

$E=\sqrt{(m{c}^{2}{)}^{2}+(pc{)}^{2}}=\sqrt{(m{c}^{2}{)}^{2}+(\gamma mvc{)}^{2}}$

with de Broglie wavelength,

$\lambda ={\displaystyle \frac{h}{p}}={\displaystyle \frac{h}{\gamma mv}}.$

I get this equation:

$E={\displaystyle \frac{h{c}^{2}}{\lambda v}}.$

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?

de Broglie Equation
Answered

Regina Ewing
2022-05-18

De Broglie relations are always written as:

$E=h\nu $

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

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
Answered

Alissa Hutchinson
2022-05-15

When I tried to derive the Wien's displacement law I used Planck's law for blackbody radiation:

${I}_{\nu}=\frac{8\pi {\nu}^{2}}{{c}^{3}}\frac{h\nu}{{e}^{h\nu /{k}_{b}T}-1}$

Asking for maximum:

$\frac{d{I}_{\nu}}{d\nu}=0:\text{}0=\frac{\mathrm{\partial}}{\mathrm{\partial}\nu}(\frac{{\nu}^{3}}{{e}^{h\nu /{k}_{b}T}-1})=\frac{3{\nu}^{2}({e}^{h\nu /{k}_{b}T}-1)-{\nu}^{3}h/{k}_{b}T\cdot {e}^{h\nu /{k}_{b}T}}{({e}^{h\nu /{k}_{b}T}-1{)}^{2}}$

It follows that numerator has to be $0$ and looking for $\nu >0$

$3({e}^{h\nu /{k}_{b}T}-1)-h\nu /{k}_{b}T\cdot {e}^{h\nu /{k}_{b}T}=0$

Solving for $\gamma =h\nu /{k}_{b}T$:

$3({e}^{\gamma}-1)-\gamma {e}^{\gamma}=0\to \gamma =2.824$

Now I look at the wavelength domain:

$\lambda =c/\nu :\text{}\lambda =\frac{hc}{\gamma {k}_{b}}\frac{1}{T}$

but from Wien's law $\lambda T=b$ I expect that $hc/\gamma {k}_{b}$ is equal to $b$ which is not:

$\frac{hc}{\gamma {k}_{b}}=0.005099$, 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:

$\frac{dI}{d\lambda}=\frac{dI}{d\nu}\frac{d\nu}{d\lambda}=\frac{c}{{\nu}^{2}}\frac{dI}{d\nu}$

where I see that $c/{\nu}^{2}$ does not influence where $d{I}_{\lambda}/d\lambda $ is zero.

Wien's displacement law
Answered

Landon Mckinney
2022-05-15

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?

Wien's displacement law
Answered

othereyeshmt4l
2022-05-15

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 Equation
Answered

revistasbibxjm87
2022-05-15

If an electron which already possesses some kinetic energy of $X\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$ is further accelerated through a potential difference of $X\prime \phantom{\rule{thinmathspace}{0ex}}\mathrm{V}$, then is it correct to state that the electron now has a total kinetic energy of $(X+X\prime )\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$? Using this information, am I allowed to substitute this value in the de Broglie equation to find the wavelength of the electron?

de Broglie Equation
Answered

Lexi Chandler
2022-05-15

De-broglie equation for uncharged particle:

$\lambda =\frac{h}{\sqrt{2mE}}$

Where, $\lambda $ = wavelength

$h$ = planks constant

$m$ = mass of uncharged particles

Wien's displacement law
Answered

deformere692qr
2022-05-14

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?

de Broglie Equation
Answered

Jace Wright
2022-05-14

In a quantum mechanics script I'm reading, the Schrödinger equation is "derived" (more precisely, motivated) by the De Broglie hypothesis. It starts at

$\lambda =\frac{2\pi h}{p}$

$\omega =\frac{E}{h}$

then takes the partial derivatives of the wave $\mathrm{\Psi}(x,t)=\mathrm{\Psi}(0,0){e}^{\frac{2\pi ix}{\lambda}-it\omega}$

$\begin{array}{}\text{(1)}& \frac{\mathrm{\partial}}{\mathrm{\partial}x}\mathrm{\Psi}(x,t)=\frac{ip}{h}\mathrm{\Psi}(x,t)\end{array}$

$\frac{\mathrm{\partial}}{\mathrm{\partial}t}\mathrm{\Psi}(x,t)=-i\frac{E}{h}\mathrm{\Psi}(x,t)$

With the non-relativistic free particle $E=\frac{1}{2m}{p}^{2}$ one gets

$\begin{array}{}\text{(2)}& ih\frac{\mathrm{\partial}}{\mathrm{\partial}t}\mathrm{\Psi}(x,t)=E\mathrm{\Psi}(x,t)=\frac{{p}^{2}}{2m}\mathrm{\Psi}(x,t)\end{array}$

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 $(\frac{\mathrm{\partial}}{\mathrm{\partial}x}\mathrm{\Psi}{)}^{2}$, which is not the second partial derivative of $\mathrm{\Psi}$ with respect to $x$.

Any hints?

de Broglie Equation
Answered

ga2t1a2dan1oj
2022-05-14

What is the validity of the relation $c=\nu \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:

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

Now, can I use $c=\nu \lambda $ to find out the frequency of the wave?

de Broglie Equation
Answered

Jaeden Weaver
2022-05-14

According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is $\lambda =h/mv$

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,

$E=h\nu ,$

where $E=$ energy, $h=$ Plank's constant ($6.62607\times {10}^{-34}\phantom{\rule{mediummathspace}{0ex}}\mathrm{J}\phantom{\rule{mediummathspace}{0ex}}\mathrm{s}$), $\nu =$ frequency.

3.Since de Broglie believes particles and wave have the same traits, the two energies would be the same:

$m{c}^{2}=h\nu .$

$m{c}^{2}=h\nu .$

4.Because real particles do not travel at the speed of light, de Broglie substituted $v$, velocity, for $c$, the speed of light:

$m{v}^{2}=h\nu .$

I want a direct proof without substituting $v$ for $c$. Is it possible to prove directly $\lambda =h/mv$ without substituting $v$ for $c$ in the equation $\lambda =h/mc$?

Wien's displacement law
Answered

Dominick Blanchard
2022-05-14

Planck's Law is commonly stated in two different ways:

${u}_{\lambda}(\lambda ,T)=\frac{2h{c}^{2}}{{\lambda}^{5}}\frac{1}{{e}^{\frac{hc}{\lambda kT}}-1}$

${u}_{\nu}(\nu ,T)=\frac{2h{\nu}^{3}}{{c}^{2}}\frac{1}{{e}^{\frac{h\nu}{kT}}-1}$

We can find the maximum of those functions by differentiating those equations with respect to $\lambda $ and to $\nu $, respectively. We get two ways to write Wien's Displacement Law:

${\lambda}_{\text{peak}}T=2.898\cdot {10}^{-3}m\cdot K$

$\frac{{\nu}_{\text{peak}}}{T}=5.879\cdot {10}^{10}Hz\cdot {K}^{-1}$

We see that ${\lambda}_{\text{peak}}\ne \frac{c}{{\nu}_{\text{peak}}}$. So what frequency or wavelength is actually detected by an optical instrument most intensely when analyzing a black body? If they are ${\lambda}_{\text{peak}}$ and ${\nu}_{\text{peak}}$, how is ${\lambda}_{\text{peak}}\ne \frac{c}{{\nu}_{\text{peak}}}$?

de Broglie Equation
Answered

Jaylene Duarte
2022-05-14

Calculate the de Broglie wavleength of a 0.05 eV ("theremal") neutron.Making a nonrelativistic calculation,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2{m}_{0}K}}={\frac{hc}{\sqrt{2\left({m}_{0}{c}^{2}\right)K}}}=\frac{12.4\times {10}^{3}\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}\cdot \text{\xc5}}{\sqrt{2(940\times {10}^{6}\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V})\left(0.05\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}\right)}}=1.28\phantom{\rule{thinmathspace}{0ex}}\text{\xc5}$

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?

Wien's displacement law
Answered

Fescoisyncsibgyp8b
2022-05-14

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).

de Broglie Equation
Answered

zuzogiecwu
2022-05-13

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
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?

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