f a photon of wavelength 0.04250 nm strikes a free electron and is scattered at an angle of ${35.0}^{\circ}$ from its original direction, find the change in the wavelength of this photon

Alexandra Richardson
2022-07-22
Answered

f a photon of wavelength 0.04250 nm strikes a free electron and is scattered at an angle of ${35.0}^{\circ}$ from its original direction, find the change in the wavelength of this photon

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Selden1f

Answered 2022-07-23
Author has **14** answers

Here, change in wavelength is determined by the relation is shown below,

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

Here,

c = velocity of light

m = mass

h= Planck's constant

By utilizing the change in wavelength and original wavelength determine the wavelength of scattered light. The difference in the energies of the incident photons and the scattered photons is equal to the change in energy of the photon. The loss in the energy of a photon is equal to the energy gained by an electron.

$\mathrm{\u25b3}\lambda =\frac{(6.626\times {10}^{-34}\text{}J\cdot s)}{(9.109\times {10}^{-31}\text{}kg)(3\times {10}^{8})}(1-\mathrm{cos}{35}^{\circ})\phantom{\rule{0ex}{0ex}}=0.43844\times {10}^{-12}\text{}m(\frac{{10}^{9}\text{}nm}{1\text{}m})\phantom{\rule{0ex}{0ex}}=4.384\times {10}^{-4}\text{}nm$

Hence, the value of change in wavelength is $4.384\times {10}^{-4}\text{}nm$

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

Here,

c = velocity of light

m = mass

h= Planck's constant

By utilizing the change in wavelength and original wavelength determine the wavelength of scattered light. The difference in the energies of the incident photons and the scattered photons is equal to the change in energy of the photon. The loss in the energy of a photon is equal to the energy gained by an electron.

$\mathrm{\u25b3}\lambda =\frac{(6.626\times {10}^{-34}\text{}J\cdot s)}{(9.109\times {10}^{-31}\text{}kg)(3\times {10}^{8})}(1-\mathrm{cos}{35}^{\circ})\phantom{\rule{0ex}{0ex}}=0.43844\times {10}^{-12}\text{}m(\frac{{10}^{9}\text{}nm}{1\text{}m})\phantom{\rule{0ex}{0ex}}=4.384\times {10}^{-4}\text{}nm$

Hence, the value of change in wavelength is $4.384\times {10}^{-4}\text{}nm$

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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.)

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.)

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What actually is giving off the light? Does the heat itself give off the light, or the metal?

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However, as I turn down the temperature, it eventually goes away completely. Is there a cut-off point for glowing?

What actually is giving off the light? Does the heat itself give off the light, or the metal?

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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?

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Why $c$ needs to be added in numerator and denominator while solving a de Broglie equation word problem?

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

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?

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?

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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?

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?

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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!

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!