Does photon behave as both a particle and a wave? Explain.

Marisol Rivers
2022-07-18
Answered

Does photon behave as both a particle and a wave? Explain.

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Tolamaes04

Answered 2022-07-19
Author has **12** answers

Light is made of small perfectly,elastic particles called photons.

Particle nature of photon :-

1. particles have momentum and kinetic energy and wavelength.

2. According to photoelectric effect when a light of certain frequency is incident on metal surface the emitted photons will have momentum ,K.E .

3. While we explain about compton effect even light consists of collection of small particles called photons

4. This shows the particle nature of photon.

Wave nature of photon :-

1. Wave will transport energy in medium and it exhibit different phenomena like interference, difraction and polarisation.

2. In youngs double slit experiment when light made to pass through the slits it forms interference pattern on screen.

3. So light is made of photons and they behave like a wave.

CONCLUSSION:- Photon has both particle and wave nature.

Particle nature of photon :-

1. particles have momentum and kinetic energy and wavelength.

2. According to photoelectric effect when a light of certain frequency is incident on metal surface the emitted photons will have momentum ,K.E .

3. While we explain about compton effect even light consists of collection of small particles called photons

4. This shows the particle nature of photon.

Wave nature of photon :-

1. Wave will transport energy in medium and it exhibit different phenomena like interference, difraction and polarisation.

2. In youngs double slit experiment when light made to pass through the slits it forms interference pattern on screen.

3. So light is made of photons and they behave like a wave.

CONCLUSSION:- Photon has both particle and wave nature.

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 2022-05-14

Proof of de Broglie wavelength for electron

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 $\lambda =h/mc$?

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 $\lambda =h/mc$?

asked 2022-04-30

De broglie equation

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.)

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.)

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 2022-05-14

What is the most intense wavelength and frequency in the spectrum of a black body?

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 ${\lambda}_{\text{peak}}\ne \frac{c}{{\nu}_{\text{peak}}}$?

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 ${\lambda}_{\text{peak}}\ne \frac{c}{{\nu}_{\text{peak}}}$?

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 2022-05-10

Wien's Displacement Law for real bodies

It is known that for perfect blackbodies,

$\lambda T=c$

where $\lambda =\text{peak wavelength}$

$T=\text{Absolute temperature}$

$c=\text{Wien's constant}$

But this is for perfect blackbodies only, which have no theoretical existence. Does a similar formula exist for real bodies, which expresses $\lambda T$ in terms of its emissitivity $\u03f5$? I googled it, but found no relevant results.

It is known that for perfect blackbodies,

$\lambda T=c$

where $\lambda =\text{peak wavelength}$

$T=\text{Absolute temperature}$

$c=\text{Wien's constant}$

But this is for perfect blackbodies only, which have no theoretical existence. Does a similar formula exist for real bodies, which expresses $\lambda T$ in terms of its emissitivity $\u03f5$? I googled it, but found no relevant results.