Finding Threshold Energy for Electrons to Produce Cerenkov Radiation

blueyellow

Homework Statement



Determine the threshold energy for electrons to produce Cerenkov radiation in glass (refractive index n=1.52).

The Attempt at a Solution



The only equations given in my notes are:

cos \theta_{C}=1/\betan for \beta>1/n

and

dN/d\lambda=\frac{2\pi\alpha}{\lambda^{2}}L sin^{2} \theta_{C}

How the second equation relates to the energy, I don't know. What does dN/d\lambda have to do with the energy? And how is the 'threshold' supposed to be worked out?

Is E=mc^{2} in this case? Is the energy supposed to be deduced from \lambda?

Please help.

My textbooks and the internet don't tell me know to do this sort of question either.
 
Physics news on Phys.org


What kind of energy exactly is that threshold referring to? Is it the kinetic energy of the electron?

Anyway, as for the problem, think about this: what do you know about Cerenkov radiation? Specifically, what does a particle need to do in order for this type of radiation to be generated?
 


The particle needs to travel faster than the local speed of light - the speed of light in the medium - glass, in this case.

So work out the speed of light in glass, and somehow link the speed of light in glass and the speed of the particle in glass. I still don't know what this has to do with the angle or 'N' though.

Thanks if anyone makes it a bit clearer.
 


blueyellow said:
The particle needs to travel faster than the local speed of light - the speed of light in the medium - glass, in this case.

So work out the speed of light in glass, and somehow link the speed of light in glass and the speed of the particle in glass.
Not "somehow" - you said exactly what the link is, that the particle has to be faster than the speed of light in glass. So what is the speed of light in glass? Can you write an inequality expressing the fact that the speed of the particle is greater than the local speed of light?

Can you then think of a way to relate that to the energy?
 


The refractive index (n) directly relates to the speed of light in the glass. n = c (speed of light in a vacuum) / v (speed of light in a medium).
 


I am rusty at this and my math is slow motion hack work. From what I read, N is the number of electrons intercepting the x axis. The angle is the angle of incidence or the angle of the cone of light emitted. I know it's an old thread, but if you have worked it, I would like to see it. That is my crude interpretation. From what I can see, it is transition radiation. I am just learning the nomenclature and trying to see simple examples at work.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top