If an incident photon of a wavelength A has scattered with an angle e after it collides with an electron in an atom, what is the shift in its wave length?

Lorena Lester
2022-07-16
Answered

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Dominique Ferrell

Answered 2022-07-17
Author has **18** answers

As per the request, I am solving the second question.

Let the photon incident on the electron (which is assumed to be at rest) has a wavelengthλ.

When the photon collides with the electrons it recoils and gets scattered at an angleθ. Let the wavelength of the scattered photon be ${\lambda}^{\prime}$. This is a classic example of an inelastic collision where an incident photon collides with an electron (assumed to be at rest) loses its energy and recoils.

Since this phenomenon is called the Compton effect.

And the shift in the wavelength is given by

$\mathrm{\u25b3}\lambda ={\lambda}^{\prime}-\lambda =\frac{h}{mc}(1-\mathrm{cos}\theta )$

Where

h is the Plank's Constant

c is the speed of light and

m is the mass of the electron.

Let the photon incident on the electron (which is assumed to be at rest) has a wavelengthλ.

When the photon collides with the electrons it recoils and gets scattered at an angleθ. Let the wavelength of the scattered photon be ${\lambda}^{\prime}$. This is a classic example of an inelastic collision where an incident photon collides with an electron (assumed to be at rest) loses its energy and recoils.

Since this phenomenon is called the Compton effect.

And the shift in the wavelength is given by

$\mathrm{\u25b3}\lambda ={\lambda}^{\prime}-\lambda =\frac{h}{mc}(1-\mathrm{cos}\theta )$

Where

h is the Plank's Constant

c is the speed of light and

m is the mass of the electron.

asked 2022-04-30

Deriving Wien's displacement law (Zettili)

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?

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?

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

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 2022-04-12

How do quasars produce light?

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?

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?

asked 2022-05-15

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?

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?

asked 2022-05-14

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?

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?

asked 2022-04-06

Is Planck temperature really the highest temperature?

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.)

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.)