Does Compton effect occur in day-to-day life too? If yes, then how can we identify it?

Alduccii2
2022-07-20
Answered

Does Compton effect occur in day-to-day life too? If yes, then how can we identify it?

You can still ask an expert for help

eishale2n

Answered 2022-07-21
Author has **15** answers

Not really. Compton scattering involves the recoil of the electron that the photon scatters from, which takes away some of the energy. This requires that the electrons are essentially free, or if they are not free that their ionization energy is much less than the energy of the incoming photon.

Visible light has insufficient energy to even ionize most materials, so under ordinary conditions it is not involved in Compton scattering.

Even if it were, shifts in wavelength due to the Compton effect are at most 5 pm. While this is easily detected with high energy photons, visible light has wavelengths of hundreds of thousands of pm, and so this effect would be very difficult to measure.

Visible light has insufficient energy to even ionize most materials, so under ordinary conditions it is not involved in Compton scattering.

Even if it were, shifts in wavelength due to the Compton effect are at most 5 pm. While this is easily detected with high energy photons, visible light has wavelengths of hundreds of thousands of pm, and so this effect would be very difficult to measure.

asked 2022-07-20

At what velocity does a proton have a 6.00-fm wavelength (about the size of a nucleus)? Assume the proton is nonrelativistic. (1 femtometer $={10}^{-15}m.$ )

asked 2022-05-08

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 $f(\mathrm{\Delta}{p}_{x},\mathrm{\Delta}{p}_{y},\mathrm{\Delta}{p}_{z},\mathrm{\Delta}x,\mathrm{\Delta}y,\mathrm{\Delta}z)\ge A$

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

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

asked 2022-05-15

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:

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

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.

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.

asked 2022-05-18

Calculating an energy of an electron with known De Broglie wavelength (why can't we calculate it similar than we do it for a photon)

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.

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.

asked 2022-05-08

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

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

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?

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?

asked 2022-05-20

Why is the de Broglie equation as well as the Schrodinger equation is correct for massive particle?

Starting from special relativity, here I see the de Broglie approximation is valid only if ${m}_{0}=0$

Derivation:

${E}^{2}={P}^{2}{C}^{2}+{m}_{0}^{2}{C}^{4}$. Here we put Plank-Einstein relation $E=h\nu =h\frac{C}{\lambda}$. Finally,

$\lambda =\frac{h}{\sqrt{{P}^{2}+{m}_{0}^{2}{C}^{2}}}\phantom{\rule{2cm}{0ex}}(1)$

If ${m}_{0}=0$ then $\lambda =\frac{h}{p}$ (de Broglie approximation).

Furthermore we know the Schrodinger equation was derived by assuming that the de Broglie approximation is true for all particles, even if ${m}_{0}\ne 0$. But if we take special relativity very strictly then this approximation looks incorrect.

In addition, if we try to derive the Schrodinger equation from the exact relation found in '1', we find completely different equation. For checking it out, lets take a wave function-

$\mathrm{\Psi}=A{e}^{i(\frac{2\pi}{\lambda}x-\omega t)}=A{e}^{i(\frac{\sqrt{{P}^{2}+{m}_{0}^{2}{C}^{2}}}{\hslash}x-\frac{E}{\hslash}t)}\phantom{\rule{2cm}{0ex}}$ (putting $\lambda $ from '1', $\frac{h}{2\pi}=\hslash $ and $E=\hslash \omega $).

Then, $\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}=-\frac{{P}^{2}+{m}_{0}^{2}{C}^{2}}{{\hslash}^{2}}\mathrm{\Psi}=-\frac{{E}^{2}}{{C}^{2}{\hslash}^{2}}\mathrm{\Psi}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}\mathrm{\Psi}=-{C}^{2}{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}\phantom{\rule{4cm}{0ex}}(2)$

Again, $\frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}=-i\frac{E}{\hslash}\mathrm{\Psi}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}E\mathrm{\Psi}=-i\hslash \frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}$

Here we see operator $E=-i\hslash \frac{\mathrm{\partial}}{\mathrm{\partial}t}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}=-{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{t}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}\mathrm{\Psi}=-{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{t}^{2}}\phantom{\rule{4cm}{0ex}}(3)$

Combining (2) and (3) we find the differential equation:

$\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}=\frac{1}{{C}^{2}}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{t}^{2}}$

It is the Maxwell's equation, not the well known Schrodiner equation!

Therefore for the Schrodinger equation to exist, the de Broglie approximation must hold for ${m}_{0}\ne 0.$ I see a clear contradiction here. Then why is the Schrodinger equation correct after all?

Starting from special relativity, here I see the de Broglie approximation is valid only if ${m}_{0}=0$

Derivation:

${E}^{2}={P}^{2}{C}^{2}+{m}_{0}^{2}{C}^{4}$. Here we put Plank-Einstein relation $E=h\nu =h\frac{C}{\lambda}$. Finally,

$\lambda =\frac{h}{\sqrt{{P}^{2}+{m}_{0}^{2}{C}^{2}}}\phantom{\rule{2cm}{0ex}}(1)$

If ${m}_{0}=0$ then $\lambda =\frac{h}{p}$ (de Broglie approximation).

Furthermore we know the Schrodinger equation was derived by assuming that the de Broglie approximation is true for all particles, even if ${m}_{0}\ne 0$. But if we take special relativity very strictly then this approximation looks incorrect.

In addition, if we try to derive the Schrodinger equation from the exact relation found in '1', we find completely different equation. For checking it out, lets take a wave function-

$\mathrm{\Psi}=A{e}^{i(\frac{2\pi}{\lambda}x-\omega t)}=A{e}^{i(\frac{\sqrt{{P}^{2}+{m}_{0}^{2}{C}^{2}}}{\hslash}x-\frac{E}{\hslash}t)}\phantom{\rule{2cm}{0ex}}$ (putting $\lambda $ from '1', $\frac{h}{2\pi}=\hslash $ and $E=\hslash \omega $).

Then, $\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}=-\frac{{P}^{2}+{m}_{0}^{2}{C}^{2}}{{\hslash}^{2}}\mathrm{\Psi}=-\frac{{E}^{2}}{{C}^{2}{\hslash}^{2}}\mathrm{\Psi}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}\mathrm{\Psi}=-{C}^{2}{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}\phantom{\rule{4cm}{0ex}}(2)$

Again, $\frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}=-i\frac{E}{\hslash}\mathrm{\Psi}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}E\mathrm{\Psi}=-i\hslash \frac{\mathrm{\partial}\mathrm{\Psi}}{\mathrm{\partial}t}$

Here we see operator $E=-i\hslash \frac{\mathrm{\partial}}{\mathrm{\partial}t}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}=-{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{t}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{2}\mathrm{\Psi}=-{\hslash}^{2}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{t}^{2}}\phantom{\rule{4cm}{0ex}}(3)$

Combining (2) and (3) we find the differential equation:

$\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{x}^{2}}=\frac{1}{{C}^{2}}\frac{{\mathrm{\partial}}^{2}\mathrm{\Psi}}{\mathrm{\partial}{t}^{2}}$

It is the Maxwell's equation, not the well known Schrodiner equation!

Therefore for the Schrodinger equation to exist, the de Broglie approximation must hold for ${m}_{0}\ne 0.$ I see a clear contradiction here. Then why is the Schrodinger equation correct after all?

asked 2022-05-18

Combination of de Broglie wavelength and mass–energy equivalence gone wrong?

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

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

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?

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?