Can a tube guru help with nu>lambda conversions?

[SPLAudioHz]

I figured this was the best place to post this thread. I am in the school of radiography and we are learning about EVERY ASPECT of tubes. Literally from the filament type to the electron production. I am sure I will need help understanding that better, but for now I need help clarifying the following:

The relationship in converting fq to wavelength (anything from angstroms to nm) then applying it to a given constant to calculate E output using a version of Planck's constant. In X-rays our version is E=12.4/lambda, 12.4 being a product of the point on the light spectrum the wavelength falls along with a few other variables that requires a bit more explanation. No need for this thread.

Is there a simpler way to convert from one to the other. I cannot seem to get it right in my head. Especially when I need to take a given kV or eV and convert the other direction.

Some of you guys are a wiz at this electronics stuff. Wondering what you do.

Thanks a boat load!!

This is for a Radiologic Physics class btw,

Jesse

[jesusdecos]

Hi Jesse,

The relation between frequency and wavelength is straightforward: nu=vp/lambda, where vp is the phase velocity of the wave, nu is the frequency of the wave and lambda is the wavelength in the medium considered. The electrical properties of the medium are given by the permittivity tensor. In an isotropic homogenuos media this is a tensor of order 0 (a scalar). In this last case, wavelength in the medium is related to wavelength in vacuum by lambda=lambda0/sqrt(epsilon_r), where lambda0 is the wavelength in vacuum and epsilon_r the relative permittivity of the medium with respect to vacuum (epsilon=epsilon_0*epsilon_r).

That is a quick overview of the theory. Now let's examine your expression. Substitute nu=vp/lambda into E=h*nu. Assuming that the medium is vacuum:
h*c = 6.626e-34 J*s * 3e8 m/s * 1eV/(1.6e-19 J) ~= 1.24e-6 eV
so here we have the expression E=1.24/lambda, with lambda in microns.

Hope you find this explanation usefull. If you have any question, please post it.
Jesús.

[SPLAudioHz]

Originally Posted by jesusdecos

Hi Jesse,

The relation between frequency and wavelength is straightforward: nu=vp/lambda, where vp is the phase velocity of the wave, nu is the frequency of the wave and lambda is the wavelength in the medium considered. The electrical properties of the medium are given by the permittivity tensor. In an isotropic homogenuos media this is a tensor of order 0 (a scalar). In this last case, wavelength in the medium is related to wavelength in vacuum by lambda=lambda0/sqrt(epsilon_r), where lambda0 is the wavelength in vacuum and epsilon_r the relative permittivity of the medium with respect to vacuum (epsilon=epsilon_0*epsilon_r).

That is a quick overview of the theory. Now let's examine your expression. Substitute nu=vp/lambda into E=h*nu. Assuming that the medium is vacuum:
h*c = 6.626e-34 J*s * 3e8 m/s * 1eV/(1.6e-19 J) ~= 1.24e-6 eV
so here we have the expression E=1.24/lambda, with lambda in microns.

Hope you find this explanation usefull. If you have any question, please post it.
Jesús.

Thank you! Yes the medium is in a vacuum. It makes more sence when you put them side by side and substitute them like that. It is always the simple things that that turn the light on for me. If I come across another question I will post it. Thanks again!

[SPLAudioHz]

Jesus,
Taking it one step further. I need help understanding something. If an amount of E is produced in the qty of 100kVp (keV, same thing) how is it determined that the fq is 2.42x10^19 Hz? If you are only given kV (keV), how does one determine fq with no other information? Is it possible?
Am I close if I were to take c (3 e8/1s) and use that as a conversion factor against given E to determine either the fq or the wavelength then solve for the remaining portion algebraically?

Thanks again,

Jesse

[jesusdecos]

Hi again. Actually, that X-Ray radiation is travelling through air. But, the relative permittivity of the air is very close to 1 and so that is its square root. The phase velocity of the wave in the air is, as you pointed out, the speed of light in vacuum. Deriving the frequency given the beam energy is straightforward:
nu = c/lambda = c*E/1.24 = 3e8 m/s * 100e3 eV / (1.24e-6 eV*m) = 2.42e19 Hz

By the way, for optical radiation and above it is common to employ either wavelength or energy rather than frequency because, as you noticed, the number is huge (i.e. we prefer refering to a 1 nanometer wave rather than a 300 PetaHertz wave).

Don't doubt to ask if you have any more questions.
Jesús.

[SPLAudioHz]

Yes your correct, I was thinking of the tube itself, not the air in which the e- travel. Wow I was on a different plain. I am still trying to convert between units of measurement in my head faster as I need to on a daily basis, but yes there are much easier ways of writing things rather than being scientifically correct. I am running out the door, but I will have one more question about how you came to 2.42e19 Hz. I see it, I can do it, just have a couple questions for better understanding.

Thanks Jesus,

Jesse