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)

Aryan Lowery
2022-10-02
Answered

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)

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Gabriella Hensley

Answered 2022-10-03
Author has **6** answers

Wavelength , $\lambda =6.6fm=6.6\times {10}^{-15}m$

To find = Velocity of proton

We can calculate the velocity using :

De Broglie wave equation :

$\lambda =\frac{h}{mv}$ , where h is Planck's constant and m is proton's mass .

Solving for v , we get :

$v=\frac{h}{m\lambda}$, where $m=1.67\times {10}^{-27}kg,h=6.626\times {10}^{-34}Js$

Substituting the given values ,we get :

$v=\frac{6.626\times {10}^{-34}}{1.67\times {10}^{-27}\times 6.6\times {10}^{-15}}$

$v=0.6\times {10}^{8}$

$v=6\times {10}^{7}$ m/s

Hence ,the velocity of the proton is $v=6\times {10}^{7}$ m/s .

To find = Velocity of proton

We can calculate the velocity using :

De Broglie wave equation :

$\lambda =\frac{h}{mv}$ , where h is Planck's constant and m is proton's mass .

Solving for v , we get :

$v=\frac{h}{m\lambda}$, where $m=1.67\times {10}^{-27}kg,h=6.626\times {10}^{-34}Js$

Substituting the given values ,we get :

$v=\frac{6.626\times {10}^{-34}}{1.67\times {10}^{-27}\times 6.6\times {10}^{-15}}$

$v=0.6\times {10}^{8}$

$v=6\times {10}^{7}$ m/s

Hence ,the velocity of the proton is $v=6\times {10}^{7}$ m/s .

asked 2022-05-08

Origin of the de Broglie Equation

I was curious about the famous $p=\hslash k$ equation. In high school I think you are just exposed to this equation with the explanation of "something something matter waves." But early in a undergraduate QM course you solve the time-independent Schrodinger's equation for a free particle in 1D and get the following solution:

$\frac{-{\hslash}^{2}}{2m}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{x}^{2}}\psi =E\psi \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\psi ={e}^{\pm ikx}$ and $E=\frac{{\hslash}^{2}{k}^{2}}{2m}$

where I believe $k$ is just defined by the $E(k)$ equation. And then do we just realize that the $E(k)$ is the classical ${p}^{2}/2m$ equation if we set $p=\hslash k$, and thus we have "derived" $p=\hslash k$ or do we "know" $p=\hslash k$ beforehand and this just confirms it?

I was curious about the famous $p=\hslash k$ equation. In high school I think you are just exposed to this equation with the explanation of "something something matter waves." But early in a undergraduate QM course you solve the time-independent Schrodinger's equation for a free particle in 1D and get the following solution:

$\frac{-{\hslash}^{2}}{2m}\frac{{\mathrm{\partial}}^{2}}{\mathrm{\partial}{x}^{2}}\psi =E\psi \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\psi ={e}^{\pm ikx}$ and $E=\frac{{\hslash}^{2}{k}^{2}}{2m}$

where I believe $k$ is just defined by the $E(k)$ equation. And then do we just realize that the $E(k)$ is the classical ${p}^{2}/2m$ equation if we set $p=\hslash k$, and thus we have "derived" $p=\hslash k$ or do we "know" $p=\hslash k$ beforehand and this just confirms it?

asked 2022-10-21

In an electron microscope, through approximately how many volts of potential difference must electrons be accelerated to achieve a de Broglie wavelength of $1.0\times {10}^{-10}m$?

a) $1.5\times {10}^{-2}V$

b) $1.5\times {10}^{-1}V$

c) $1.5V$

d) $15V$

e) $150V$

a) $1.5\times {10}^{-2}V$

b) $1.5\times {10}^{-1}V$

c) $1.5V$

d) $15V$

e) $150V$

asked 2022-08-13

How do you calculate the velocity of an electron if the debroy wavelength is given.

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 2022-05-13

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 2022-10-19

Please explain in detail with steps how to set up and solve for this equation.

Find the de Broglie wavelength for a 1500kg car when its speed is 80km/h. How significant are the wave properties of this car likely to be?

Find the de Broglie wavelength for a 1500kg car when its speed is 80km/h. How significant are the wave properties of this car likely to be?

asked 2022-04-27

De Broglie's Matter wave equation dividing by zero

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?

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?