The speed of electron.

Israel Hale
2022-07-20
Answered

The speed of electron.

You can still ask an expert for help

kamphundg4

Answered 2022-07-21
Author has **20** answers

Explanation of Solution

Using De Broglie’s equation,

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

Here, $\lambda $ is the wavelength, h is the Planck’s constant, m is the mass of electron and v is the speed of electron.

Re-arrange the equation to get v.

$v=\frac{h}{m\lambda}$

Conclusion:

Substitute $6.63\times {10}^{-34}J\cdot s$ for h, $9.11\times {10}^{-31}kg$ for m and 0.1 nm for $\lambda $ to get v

$v=\frac{6.63\times {10}^{-34}J\cdot s}{(9.11\times {10}^{-31}kg)(0.1\hspace{0.17em}nm)}$

$=\frac{6.63\times {10}^{-34}J\cdot s}{(9.11\times {10}^{-31}kg)(0.1\times {10}^{-9}m)}$

$=7.28\times {10}^{6}m/s$

Therefore, speed of electron is $=7.28\times {10}^{6}m/s$

Using De Broglie’s equation,

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

Here, $\lambda $ is the wavelength, h is the Planck’s constant, m is the mass of electron and v is the speed of electron.

Re-arrange the equation to get v.

$v=\frac{h}{m\lambda}$

Conclusion:

Substitute $6.63\times {10}^{-34}J\cdot s$ for h, $9.11\times {10}^{-31}kg$ for m and 0.1 nm for $\lambda $ to get v

$v=\frac{6.63\times {10}^{-34}J\cdot s}{(9.11\times {10}^{-31}kg)(0.1\hspace{0.17em}nm)}$

$=\frac{6.63\times {10}^{-34}J\cdot s}{(9.11\times {10}^{-31}kg)(0.1\times {10}^{-9}m)}$

$=7.28\times {10}^{6}m/s$

Therefore, speed of electron is $=7.28\times {10}^{6}m/s$

asked 2022-05-08

Importance of Schrodinger equation

Louis de Broglie suggested that for microparticles like electrons, wave-like properties can be applied in order to explain some phenomena. Schrodinger wrote down an equation, a wave equation, describing these waves. What I do not understand is why is Schrodinger's contribution so important; if the concept of wave-like property of an electron was known already, then why writing a mere wave equation was an important step?

Louis de Broglie suggested that for microparticles like electrons, wave-like properties can be applied in order to explain some phenomena. Schrodinger wrote down an equation, a wave equation, describing these waves. What I do not understand is why is Schrodinger's contribution so important; if the concept of wave-like property of an electron was known already, then why writing a mere wave equation was an important step?

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 2022-05-14

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?

asked 2022-05-10

Which formula for the de Broglie wavelength of an electron is correct?

So, I have my exams in physics in a week, and upon reviewing I was confused by the explanation of de Broglie wavelength of electrons in my book. Firstly, they stated that the equation was: $\lambda =\frac{h}{p}$, where $h$ is the Planck constant, and $p$ is the momentum of the particle. Later, however, when talking about electron diffraction and finding the angles of the minima, the author gave the formula equivalent to that for light: $\lambda =\frac{hc}{E}$. Now, what I don't understand is if it is simply a mistake made by the author, or whether a different formula have to be used for electron diffraction, as the two formulae are very clearly not equivalent. In the latter case, I don't understand why the formula would be different. I greatly appreciate the help, as the exams are really close, and I would like to make sure I get this right!

Edit: I was told that pictures of text are taking away from the readability of the posts, and thus they were removed. Essentially, the difference between the two cases are that in the first case, the proton did not have any significantly large kinetic energy, while in the second example, the kinetic energy was $400\text{}MeV$

So, I have my exams in physics in a week, and upon reviewing I was confused by the explanation of de Broglie wavelength of electrons in my book. Firstly, they stated that the equation was: $\lambda =\frac{h}{p}$, where $h$ is the Planck constant, and $p$ is the momentum of the particle. Later, however, when talking about electron diffraction and finding the angles of the minima, the author gave the formula equivalent to that for light: $\lambda =\frac{hc}{E}$. Now, what I don't understand is if it is simply a mistake made by the author, or whether a different formula have to be used for electron diffraction, as the two formulae are very clearly not equivalent. In the latter case, I don't understand why the formula would be different. I greatly appreciate the help, as the exams are really close, and I would like to make sure I get this right!

Edit: I was told that pictures of text are taking away from the readability of the posts, and thus they were removed. Essentially, the difference between the two cases are that in the first case, the proton did not have any significantly large kinetic energy, while in the second example, the kinetic energy was $400\text{}MeV$

asked 2022-05-08

If a CMB photon traveled for 13.7 billion years (- 374,000 years) to reach me. How far away was the source of that CMB photon when it first emitted it?

My attempt to solve this question was to use the following assumptions:

1.Temperature of CMB photon today is 2.725 K (will use value of 3 K here)

2.Temperature of CMB photon when it was first emitted is 3000 K

3.A factor of x1000 in temperature decrease results in a factor of x1000 in wavelength increase. (According to Wien's displacement law)

Does this mean that the source of the CMB photon that just reached me today, was actually 13.7 billion light years / 1000 = 13.7 million light years away from me when it first emitted the photon?

My attempt to solve this question was to use the following assumptions:

1.Temperature of CMB photon today is 2.725 K (will use value of 3 K here)

2.Temperature of CMB photon when it was first emitted is 3000 K

3.A factor of x1000 in temperature decrease results in a factor of x1000 in wavelength increase. (According to Wien's displacement law)

Does this mean that the source of the CMB photon that just reached me today, was actually 13.7 billion light years / 1000 = 13.7 million light years away from me when it first emitted the photon?

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 2022-05-08

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!