To determine

To Calculate: Momentum of photon by using equation $p=mv$

To Calculate: Momentum of photon by using equation $p=mv$

Krish Logan
2022-10-14
Answered

To determine

To Calculate: Momentum of photon by using equation $p=mv$

To Calculate: Momentum of photon by using equation $p=mv$

You can still ask an expert for help

Layton Leach

Answered 2022-10-15
Author has **15** answers

Light is composed of photons which do not have any rest mass. So, one cannot find the momentum of a photon using the equation $p=mv$ , which requires mass for the calculation of the momentum.

The momentum of a photon can be calculated using the de Broglie wavelength equation:

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

Where, h is the Planck’s constant and $\lambda $ is the wavelength.

Conclusion:

One cannot use $p=mv$ to find the momentum of a photon.

The momentum of a photon can be calculated using the de Broglie wavelength equation:

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

Where, h is the Planck’s constant and $\lambda $ is the wavelength.

Conclusion:

One cannot use $p=mv$ to find the momentum of a photon.

asked 2022-05-07

Direction of momentum given by the de Broglie relation

$p=mv$

where $m$ is the mass of an electron, and $v$ is its velocity. In this case, since $v$ is a vector, it's clear that the momentum will be also a vector.

However if the momentum is a vector quantity (and it is), what is the direction of the electron's momentum given by the de Broglie relation

$p=h/\lambda \phantom{\rule{0ex}{0ex}}p=\hslash k$

if the Planck constant $h$ is scalar and the wavelength $\lambda $ is also scalar. Similarly the reduced Planck constant $\hslash $ is scalar and the wavenumber $k=2\pi /\lambda $ is also scalar.

$p=mv$

where $m$ is the mass of an electron, and $v$ is its velocity. In this case, since $v$ is a vector, it's clear that the momentum will be also a vector.

However if the momentum is a vector quantity (and it is), what is the direction of the electron's momentum given by the de Broglie relation

$p=h/\lambda \phantom{\rule{0ex}{0ex}}p=\hslash k$

if the Planck constant $h$ is scalar and the wavelength $\lambda $ is also scalar. Similarly the reduced Planck constant $\hslash $ is scalar and the wavenumber $k=2\pi /\lambda $ is also scalar.

asked 2022-05-08

De Broglie Wavelength interpretation

I've just started learning about the double slit experiment (just in the short appendix section in Schroeder's Thermal Physics), and I'm extremely confused by this one thing:

In it, out of basically nowhere he pulls out the De Broglie equation, that λ = h/p.

I've studied double slit diffraction before, and I've been trying to connect them in order to understand what this wavelength actually means.

In double slit diffraction, when the wavelength is larger, the diffraction "stripes" that form on the wall appear further apart. They also appear larger.

y = $\frac{m\lambda L}{d}$ (approx, considering the distance to the screen is really large and thus almost parallel rays (drawn out waves) can have a path difference and interfere)

If we were to make the wavelength extremely small, that would mean that anything a little off-center would interfere, so the smaller the wavelength, the closer together the "stripes" on the wall would be.

Now, when we connect the 2 equations, this means that the faster the electrons are moving (the smaller their wavelength) the more places they will interfere on the wall, and therefore there will be a lesser distance between adjacent places where the electrons hit (bright spots) and places where they don't (dark spots).

The way I'm interpreting this is that the smaller the "wavelength" of the electron, the more the probability it has to have been in different places at the same time, that is, the less we can know its position. That's why more stripes will appear on the detecting screen because there are more positions which the electron could've been in, and since its technically in all of them at the same time while it travels, it can interfere with itself more.

Is this interpretation correct? Does a faster momentum (a smaller wavelength) mean that the electron literally is at more places at the same time while it travels from the electron gun, through the slits, and to the wall? Thank you!

I've just started learning about the double slit experiment (just in the short appendix section in Schroeder's Thermal Physics), and I'm extremely confused by this one thing:

In it, out of basically nowhere he pulls out the De Broglie equation, that λ = h/p.

I've studied double slit diffraction before, and I've been trying to connect them in order to understand what this wavelength actually means.

In double slit diffraction, when the wavelength is larger, the diffraction "stripes" that form on the wall appear further apart. They also appear larger.

y = $\frac{m\lambda L}{d}$ (approx, considering the distance to the screen is really large and thus almost parallel rays (drawn out waves) can have a path difference and interfere)

If we were to make the wavelength extremely small, that would mean that anything a little off-center would interfere, so the smaller the wavelength, the closer together the "stripes" on the wall would be.

Now, when we connect the 2 equations, this means that the faster the electrons are moving (the smaller their wavelength) the more places they will interfere on the wall, and therefore there will be a lesser distance between adjacent places where the electrons hit (bright spots) and places where they don't (dark spots).

The way I'm interpreting this is that the smaller the "wavelength" of the electron, the more the probability it has to have been in different places at the same time, that is, the less we can know its position. That's why more stripes will appear on the detecting screen because there are more positions which the electron could've been in, and since its technically in all of them at the same time while it travels, it can interfere with itself more.

Is this interpretation correct? Does a faster momentum (a smaller wavelength) mean that the electron literally is at more places at the same time while it travels from the electron gun, through the slits, and to the wall? Thank you!

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 2022-05-13

Special Relativity, Louis de Broglie Equation Dilemma

While learning atomic structure I stumbled upon a very unusual doubt.

As we know that the energy of a wave is given by the equation: $E=\frac{hc}{\lambda}$ and Louis de broglie wave equation is given by the equation ${\lambda}_{B}=\frac{h}{p}$. My doubt is that, that is ${\lambda}_{B}=\lambda $. Do the ${\lambda}_{B},\lambda $ represent the same thing $?$

My teacher equated $E=\frac{hc}{\lambda}$ and $E=mc\xb2$ to form $\frac{hc}{\lambda}=mc\xb2$ and rearranged to form $\lambda =\frac{h}{mc}$ and then replaced $\lambda $ by ${\lambda}_{{\rm B}}$ and $c$ by $v$ for general formula and derived the Louis de broglie equation. This created my doubt in first place and I created another doubt that whether the equation for energy of wave is valid for relativistic equation of $E=mc\xb2$ because the $E=mc\xb2$ is for particles while the former is for waves.

Is my understanding correct$?$ Please help and thanks in advance$!$

While learning atomic structure I stumbled upon a very unusual doubt.

As we know that the energy of a wave is given by the equation: $E=\frac{hc}{\lambda}$ and Louis de broglie wave equation is given by the equation ${\lambda}_{B}=\frac{h}{p}$. My doubt is that, that is ${\lambda}_{B}=\lambda $. Do the ${\lambda}_{B},\lambda $ represent the same thing $?$

My teacher equated $E=\frac{hc}{\lambda}$ and $E=mc\xb2$ to form $\frac{hc}{\lambda}=mc\xb2$ and rearranged to form $\lambda =\frac{h}{mc}$ and then replaced $\lambda $ by ${\lambda}_{{\rm B}}$ and $c$ by $v$ for general formula and derived the Louis de broglie equation. This created my doubt in first place and I created another doubt that whether the equation for energy of wave is valid for relativistic equation of $E=mc\xb2$ because the $E=mc\xb2$ is for particles while the former is for waves.

Is my understanding correct$?$ Please help and thanks in advance$!$

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 2022-05-15

De Broglie wavelength of an electron

If an electron which already possesses some kinetic energy of $X\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$ is further accelerated through a potential difference of $X\prime \phantom{\rule{thinmathspace}{0ex}}\mathrm{V}$, then is it correct to state that the electron now has a total kinetic energy of $(X+X\prime )\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$? Using this information, am I allowed to substitute this value in the de Broglie equation to find the wavelength of the electron?

If an electron which already possesses some kinetic energy of $X\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$ is further accelerated through a potential difference of $X\prime \phantom{\rule{thinmathspace}{0ex}}\mathrm{V}$, then is it correct to state that the electron now has a total kinetic energy of $(X+X\prime )\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{V}$? Using this information, am I allowed to substitute this value in the de Broglie equation to find the wavelength of the electron?

asked 2022-05-14

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

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?

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?