At what velocity does a proton have a 6.00-fm wavelength (about the size of a nucleus)? Assume the proton is nonrelativistic. (1 femtometer $={10}^{-15}m.$ )

makaunawal5
2022-07-20
Answered

At what velocity does a proton have a 6.00-fm wavelength (about the size of a nucleus)? Assume the proton is nonrelativistic. (1 femtometer $={10}^{-15}m.$ )

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emerhelienapj

Answered 2022-07-21
Author has **14** answers

The de Broglie equation is,

$v=\frac{h}{{m}_{\text{proton}}\lambda}$

Therefore,

$v=\frac{6.62\times {10}^{-34}kg{m}^{2}/s}{(1.67\times {10}^{-27}kg)(6.00\times {10}^{-15}m)}\phantom{\rule{0ex}{0ex}}=6.61\times {10}^{7}m/s$

$v=\frac{h}{{m}_{\text{proton}}\lambda}$

Therefore,

$v=\frac{6.62\times {10}^{-34}kg{m}^{2}/s}{(1.67\times {10}^{-27}kg)(6.00\times {10}^{-15}m)}\phantom{\rule{0ex}{0ex}}=6.61\times {10}^{7}m/s$

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?

asked 2022-05-13

Wavelength and relativity

From de Broglie equation λ=h/p. But p=mv and velocity is a relativistic quantity so also wavelength is relative ? In other words does wavelength depends on the reference frame ?

From de Broglie equation λ=h/p. But p=mv and velocity is a relativistic quantity so also wavelength is relative ? In other words does wavelength depends on the reference frame ?

asked 2022-05-15

Deeper underlying explanation for color-shift in Wien's Law?

Suppose we have a blackbody object, maybe a star or a metal (although I understand neither of these are actually blackbody objects, to some extent my understanding is they can approximate one). According to Wien's Law, as temperature increases, peak wavelength will decrease, so the color we observe will "blueshift." Despite having reached Wien's Law in my research, my understanding is that this is in fact not an answer to why the blueshift occurs, but just a statement that the blueshift does occur.

So, is there some deeper reason why blackbodies peak wavelength emitted decreases with temperature?

Suppose we have a blackbody object, maybe a star or a metal (although I understand neither of these are actually blackbody objects, to some extent my understanding is they can approximate one). According to Wien's Law, as temperature increases, peak wavelength will decrease, so the color we observe will "blueshift." Despite having reached Wien's Law in my research, my understanding is that this is in fact not an answer to why the blueshift occurs, but just a statement that the blueshift does occur.

So, is there some deeper reason why blackbodies peak wavelength emitted decreases with temperature?

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Showing Wien's Displacement Law from Wien's Law

Does anyone know how I would show that $\lambda \ast T$ is constant, using only Wien's Law? That $\rho (\lambda ,T)=1/{\lambda}^{5}\ast f(\lambda T)$ I differentiated, but all I could get was $\lambda T=5f(\lambda T)/{f}^{\prime}(\lambda T)$, which I don't think means it's necessarily a constant.

Does anyone know how I would show that $\lambda \ast T$ is constant, using only Wien's Law? That $\rho (\lambda ,T)=1/{\lambda}^{5}\ast f(\lambda T)$ I differentiated, but all I could get was $\lambda T=5f(\lambda T)/{f}^{\prime}(\lambda T)$, which I don't think means it's necessarily a constant.

asked 2022-05-18

What causes hot things to glow, and at what temperature?

I have an electric stove, and when I turn it on and turn off the lights, I notice the stove glowing.

However, as I turn down the temperature, it eventually goes away completely. Is there a cut-off point for glowing?

What actually is giving off the light? Does the heat itself give off the light, or the metal?

I have an electric stove, and when I turn it on and turn off the lights, I notice the stove glowing.

However, as I turn down the temperature, it eventually goes away completely. Is there a cut-off point for glowing?

What actually is giving off the light? Does the heat itself give off the light, or the metal?

asked 2022-05-14

Is the relation c=νλ valid only for Electromagnetic waves?

What is the validity of the relation $c=\nu \lambda $? More specifically, is this equation valid only for Electromagnetic waves?

I read this statement in a book, which says:

"de Broglie waves are not electromagnetic in nature, because they do not arise out of accelerated charged particle."

This seems correct, but arises a doubt in my mind.

Suppose I find out the wavelength of a matter wave (or de Broglie wave) using de Broglie's wave equation:

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

Now, can I use $c=\nu \lambda $ to find out the frequency of the wave?

What is the validity of the relation $c=\nu \lambda $? More specifically, is this equation valid only for Electromagnetic waves?

I read this statement in a book, which says:

"de Broglie waves are not electromagnetic in nature, because they do not arise out of accelerated charged particle."

This seems correct, but arises a doubt in my mind.

Suppose I find out the wavelength of a matter wave (or de Broglie wave) using de Broglie's wave equation:

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

Now, can I use $c=\nu \lambda $ to find out the frequency of the wave?

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.