Calculate the de Broglie wavelength of electrons accelerated through 3868 V. Round off the answer to 2 decimal places with scientific representation.

Riya Andrews
2022-10-09
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Krha77

Answered 2022-10-10
Author has **8** answers

Given data: -

The potential difference is V = 3868 V.

Here, the mass of the electron is $m=9.1\times {10}^{-31}kg$, the value of the Planck's constant is $h=6.62\times {10}^{-34}$ J-s and the charge on electron is $e=1.6\times {10}^{-19}$ C.

The expression for the DE Broglie wavelength of electrons is given as,

$\lambda =\frac{h}{\sqrt{2meV}}$

Substitute all the known values in above equation,

$\lambda =\frac{6.62\times {10}^{-34}J\cdot s}{\sqrt{2\times 9.1\times {10}^{-31}kg\times 1.6\times {10}^{-19}C\times 3868V}}\phantom{\rule{0ex}{0ex}}\lambda =1.97\times {10}^{-11}m$

Thus, the DE Broglie wavelength of electrons is $\lambda =1.97\times {10}^{-11}m$

The potential difference is V = 3868 V.

Here, the mass of the electron is $m=9.1\times {10}^{-31}kg$, the value of the Planck's constant is $h=6.62\times {10}^{-34}$ J-s and the charge on electron is $e=1.6\times {10}^{-19}$ C.

The expression for the DE Broglie wavelength of electrons is given as,

$\lambda =\frac{h}{\sqrt{2meV}}$

Substitute all the known values in above equation,

$\lambda =\frac{6.62\times {10}^{-34}J\cdot s}{\sqrt{2\times 9.1\times {10}^{-31}kg\times 1.6\times {10}^{-19}C\times 3868V}}\phantom{\rule{0ex}{0ex}}\lambda =1.97\times {10}^{-11}m$

Thus, the DE Broglie wavelength of electrons is $\lambda =1.97\times {10}^{-11}m$

asked 2022-05-13

Blackbody radiation, de Broglie equation, and lightwaves to be shifted left

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!

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 2022-10-20

In de Broglie's equation w=wavelength, $w=\frac{h}{mv}$. How do I rearrange to solve for mass?

asked 2022-10-02

Consider a proton with a 6.6 fm wavelength. What is the velocity of the proton in meters per second? Assume the proton is nonrelativistic. (1 femtometer $={10}^{-15}$ m)

asked 2022-05-18

Are De Broglie relations only applicable to particles that have zero potential energy?

De Broglie relations are always written as:

$E=h\nu $

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

However, it doesn't make sense when you have waves that are eigenstates of a Hamiltonian operator with a non-zero potential. That is because that would give us a direct relation between momentum and energy: $\frac{E}{p}={v}_{p}$, where ${v}_{p}$ is the phase velocity. This doesn't seem to make sense to me. So, would these equations be the same with the electron of the hydrogen atom, for example?

De Broglie relations are always written as:

$E=h\nu $

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

However, it doesn't make sense when you have waves that are eigenstates of a Hamiltonian operator with a non-zero potential. That is because that would give us a direct relation between momentum and energy: $\frac{E}{p}={v}_{p}$, where ${v}_{p}$ is the phase velocity. This doesn't seem to make sense to me. So, would these equations be the same with the electron of the hydrogen atom, for example?

asked 2022-07-22

John Isner holds the ATP's (Association of Tennis Professionals) official record for the fastest serve at 253 km/h. Calculate the de Broglie wavelength of a tennis ball with the standard weight of 58.8 grams. Give your answer in ${10}^{-34}m$ (for example, if your answer is $5.6\times {10}^{-34}$ m, then enter 5.6). This should give you a good idea of why one cannot expect to observe interference affects using tennis balls rather then, for example, electrons.

asked 2022-08-06

Find the de Broglie wavelength for a baseball weighing 149g that is moving at 101mph (fastball).