Is there a limit on how much energy a photon might have in a FOR?

[Aurelius120]

Homework Statement

Kinetic Energy of an electron is given by $$\frac{hc}{\lambda_{ph}}-\phi+eV$$
Does $V>>>\frac{\phi}{e}$ mean that kinectic energy is approximately $eV$?

Relevant Equations

NA

So there was this question:

The first option seems to be the only correct answer.
$$\lambda_e=\dfrac{h}{\sqrt{2m(KE)}}$$. The answer would be correct if $KE \approx eV$
The option mentions that $eV>>\phi$ so $\phi$ can be ignored.

But I don't think that necessarily means that the energy of photon can be ignored?

So I looked it up and this seems to conclude that there is no limit on energy of photon. This would mean that the option 1 will be correct only if $eV>>\dfrac{hc}{\lambda_{ph}}$ (which is not specified so incorrect.) Can we reasonably assume that for most practical purposes, the energy of photon is of same order as work function($\phi$)?

 

[anuttarasammyak]

"Minimum de Broglie length", the problem statement says. It is the case
\phi=h c / \lambda_{ph}

 

[Steve4Physics]

Homework Statement: Kinetic Energy of an electron is given by $$\frac{hc}{\lambda_{ph}}-\phi+eV$$
Does $V>>>\frac{\phi}{e}$ mean that kinectic energy is approximately $eV$?
Relevant Equations: NA

So there was this question:
View attachment 347873

The first option seems to be the only correct answer.
$$\lambda_e=\dfrac{h}{\sqrt{2m(KE)}}$$. The answer would be correct if $KE \approx eV$
The option mentions that $eV>>\phi$ so $\phi$ can be ignored.

But I don't think that necessarily means that the energy of photon can be ignored?

So I looked it up and this seems to conclude that there is no limit on energy of photon. This would mean that the option 1 will be correct only if $eV>>\dfrac{hc}{\lambda_{ph}}$ (which is not specified so incorrect.) Can we reasonably assume that for most practical purposes, the energy of photon is of same order as work function($\phi$)?

The question specifically refers to light which has photon energies in the range 1.6eV to 3.3eV approx.

Note, ‘eV’ (electron-volt) is an energy unit, not to be confused with the ‘$eV$’ in your equations. To avoid confusion I’ll use $V_a$ to represent the accelerating voltage.

The electron’s final energy ($K$) will be the sum of:
- the photon energy (a few eV);
- minus the work function (typically a few eV);
- the energy gained due to the accelerating voltage ($eV_a$).

Clearly if $V_a$ is large enough, $K$ is nearly equal to $eV_a$ because the other terms will be negligible.

However, it would have been better if the question had said $V \gg \frac 1e (\frac {hc}{\lambda_{ph}}-\phi)$.

 

[kuruman]

In the title, what does "FOR" stand for?

 

[Aurelius120]

In the title, what does "FOR" stand for?

Frame Of Reference
In that stackexchange post, they said they can increase the photons energy by changing the frame of reference

 

[Aurelius120]

"Minimum de Broglie length", the problem statement says. It is the case
\phi=h c / \lambda_{ph}

Shouldn't that mean maximum kinetic energy therefore as high as possible value of photon energy?

 

[Aurelius120]

The question specifically refers to light which has photon energies in the range 1.6eV to 3.3eV approx.

Oh ! Light means visible light?
I thought all EM waves can be called types of light.
UV light, IR light, etc.

In hindsight Gamma light, Microwave Light, Radio wave Light sound odd
X-Ray light sounds alright.

Even then is the answer to title question yes?

 

[jbriggs444]

Even then is the answer to title question yes?

Is there a limit to the energy of a photon if we are free to change to an arbitrary frame of reference?

No. There is no such limit. We can pick a frame moving arbitrarily close to the speed of light (relative to some initial frame).

This post by @jtbell in 2010 gives the Lorentz transform for energy and momentum under such a boost.

 

[Aurelius120]

Is there a limit to the energy of a photon if we are free to change to an arbitrary frame of reference?

No. There is no such limit. We can pick a frame moving arbitrarily close to the speed of light (relative to some initial frame).

This post by @jtbell in 2010 gives the Lorentz transform for energy and momentum under such a boost.

So is there a limit if the frame of reference is fixed?

 

[jbriggs444]

So is there a limit if the frame of reference is fixed?

If we select co-moving coordinates then the total energy in the observable universe is an upper bound on photon energy. Not a particularly tight bound -- the ability to collect that energy into a single photon would pose a significant engineering challenge.

 

[Steve4Physics]

In hindsight Gamma light, Microwave Light, Radio wave Light sound odd
X-Ray light sounds alright.

I'd agree. The term 'light' is typically applied to IR, visible light , UV and (sometimes) X-rays.

But from the nature of the question, it would be reasonable to assume that 'light' refers to visible light, 'near UV' and possibly 'near IR'. I'm sure that's' what the author of the question intended.

Note that for sufficiently high energy photons, the photoelectric effect would be insignificant. Other processes (notably Compton scattering and pair-production) would occur. This is outside the scope of the question.