Recent questions in Quantum Mechanics

de Broglie Equation
Answered

znacimavjo
2022-05-03

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, $E=hf$

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

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

$E=m{c}^{2}$

$E=\frac{p}{c}\cdot {c}^{2}$

$E=\frac{hc}{\lambda}=hf$

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?

Wien's displacement law
Answered

acceloq3c
2022-04-30

According to Zettili, we can derive Wien's displacement law from Planck's energy density

$\stackrel{~}{u}(\lambda ,T)=\frac{8\pi hc}{{\lambda}^{5}}\frac{1}{{e}^{hc/kT\lambda}-1}$

where $\lambda $ is the wavelength and $T$ is the temperature. Taking the derivative of this, setting it equal to zero, and rearranging terms, one gets

$\frac{\alpha}{\lambda}=5(1-{e}^{-\alpha /\lambda})$

where $\alpha =hc/(kT)$. To solve this, Zettili goes on to say that we can write

$\frac{\alpha}{\lambda}=5-\u03f5$

so that the previously mentioned equation becomes

$5-\u03f5=5-5{e}^{-5+\u03f5}$

Apparently, $\u03f5\approx 5{e}^{-5}$ is a good answer. My first question is why do we use $5-\u03f5$ 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 $\u03f5\approx 5{e}^{-5}$ when there is an extra factor of ${e}^{\u03f5}$ in the equation? Why can we ignore that factor?

de Broglie Equation
Answered

Davin Sheppard
2022-04-30

What is the de Broglie wavelength? Also, does the $\lambda $ sign in the de Broglie equation stand for the normal wavelength or the de Broglie wavelength? If $\lambda $ is the normal wavelength of a photon or particle, is $\lambda \propto \frac{1}{m}$ 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.)

Wien's displacement law
Answered

Caitlyn Cole
2022-04-30

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.

Wien's displacement law
Answered

Ezequiel Olson
2022-04-28

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 ${E}_{photon}=h\cdot f$. I do know that classically, the energy of an oscillator is $E={K}_{b}\cdot T$. Is it okay to use this formula for this question?

de Broglie Equation
Answered

pettingyg0
2022-04-27

I was just thinking about De Broglie's matter wave equation: $\lambda =\frac{h}{p}$ where $p$ is the momentum of the object. But what if the object is at rest? Won't we be dividing by zero? What if we take the limit as momentum tends to zero, won't we start to get noticeable waves? Can someone please explain to me where I went wrong?

Wien's displacement law
Answered

lnwlf1728112xo85f
2022-04-12

So I'm reading an introductory book on Quantum Theory (David Park, 3rd ed.) and I am having trouble with the following question:

"According to Wien's displacement law, the wavelength ${\lambda}_{m}$ at which blackbody radiation at temperature T has its maximum intensity is given roughly by ${\lambda}_{m}T\simeq $ 3 mm K. Assuming that the quantum energy at this temperature is of the order of kT where k is Boltzmann's constant, estimate the value of Planck's constant."

My attempt at a solution is as follows:

${\lambda}_{m}T\simeq $ 3 mm K $,\phantom{\rule{1em}{0ex}}$$\lambda =\frac{c}{\nu}$$\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}$$\frac{cT}{\nu}=3\cdot {10}^{-3}mK$$\cdot (\frac{k}{c})\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}$$\frac{kT}{\nu}=1.38\cdot {10}^{-34}Js$

$(\frac{1}{2}m{v}^{2}{)}_{max}=k{T}_{max}=h\nu -e\varphi $

$\frac{k{T}_{max}}{\nu}+\frac{e\varphi}{\nu}=h$

$1.38\cdot {10}^{-34}+\frac{e\varphi}{\nu}=h$

At this point I get stuck, the orders of magnitude and units seem to be right but I'm not sure what to invoke/what is given in the question that can solve/eliminate this last term to get h, or if I'm even on the right track.

Wien's displacement law
Answered

Stoyanovahvsbh
2022-04-12

Wikipedia states that the radiation of quasars is partially nonthermal (i.e. not blackbody).

Well, what percentage of the total radiation isn't black body?

For some reason Wikipedia doesn't give a number for the temperature of the gas of the accretion disc of a quasar. I had to search for a while and tried many keywords on Google to get an answer of 80 million Kelvin. If someone knows better and/or can give me a range instead of a single number that would be nice.

Based on that number and an online calculator of wien's displacement law I got that quasars emit blackbody radiation mainly on the x-ray and gamma ray section of the spectrum. Is that true?

Wien's displacement law
Answered

indimiamimactjcf
2022-04-12

I have two questions related to inferring properties of stars from their masses and radii.

1.What properties of a star's spectrum could we deduce? In particular, do all stars emit like black bodies? If so, I'm guessing that one can calculate the star's effective temperature and use Wien's displacement law to find the peak of the black body spectrum.

2.How would the mass and radius influence Type 1 X-ray bursts?

Wien's displacement law
Answered

lifretatox8n
2022-04-12

The thing about infrared thermometers that bugs me is how can you get the same temperature reading regardless of the distance to the object. Shouldn't there be a difference when measuring from two different standing points since energy flux density decreases with $\frac{1}{distanc{e}^{2}}$ and infrared thermometers work by focusing IR light on a thermopile, which then results in decreased (when measuring from further away) absorbed energy and therefore lower temperature and finally lower voltage across thermopile. Is there something I am getting wrong about this, or do IR thermometers make use of some other physics law like Wien's displacement law, by somehow measuring ${\lambda}_{peak}$ to determine the temperature?

Wien's displacement law
Answered

Alisa Durham
2022-04-12

If I calculate the temperature of the Sun by replacing the effective frequency of the Sun (598 THz) in Planck's curve for Blackbody radiation, the result will be 10170 K. Still, in Wien's displacement law, the result will be 5778 K, is there an answer for this problem?

Wien's displacement law
Answered

encamineu2cki
2022-04-12

Does anyone know how I would show that $\lambda \ast T$ is constant, using only Wien's Law? That $\rho (\lambda ,T)=1/{\lambda}^{5}\ast f(\lambda T)$ I differentiated, but all I could get was $\lambda T=5f(\lambda T)/{f}^{\prime}(\lambda T)$, which I don't think means it's necessarily a constant.

Wien's displacement law
Answered

Jayla Faulkner
2022-04-06

Actually I was learning about Wien's displacement law. It states that

$\lambda T=2.898\times {10}^{-3}mK$

This is actually a part of Planck's law where the Planck's constant originated.

Now Planck's temperature is given as

${T}_{p}=\sqrt{\frac{h{c}^{5}}{2\pi G{k}_{b}^{2}}}=1.416\times {10}^{32}K$

Now Planck's length is $1.616\times {10}^{-35}m$

Now since the smallest possible wavelength is Planck's length, we can say wavelength of the electromagnetic radiation is Planck's length (Assume the energy doesn't create a black hole).

Now according to Wien's displacement law,

${l}_{p}T=2.898\times {10}^{-3}mK$

Now solving this we get $1.79\times {10}^{32}K$, which is higher than the actual Planck's temperature.

Since this displacement law is completely derived from Planck's law, it bit frustrated me. I'm a bit confused. Is it the limit of the displacement law or my flaw?

Please rectify this.

(Sorry if I made any mistake. I'm new to this one. Please explain my mistake. I'm glad to hear that.)

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