# Calculation of the least value of PD needed to produce x-rays

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1. Apr 4, 2017

1. The problem statement, all variables and given/known data

In my excercise book, I was given a typical diagram as that in Fig (6-7) above.
And the problem statement says:
The diagram presents the x-ray spectrum produced by Coolidge tube. Determine the least potential difference needed to produce the characteristic radiation.

2. Relevant equations
E= hƒ = hc/λ where: h is Planck's constant, ƒ is the frequency, c is light speed and λ is the wavelength)
K.E= eV

3. The attempt at a solution

I really don't know how to start in this, because the diagram doesn't appear to be precise, I mean I don't know actually the precise shortest wavelength where the characteristic radiation appears in the diagram, would it be 0.07nm?!

If it would be 0.07nm, I will do this:

E=hc/λ=(6.625*10^-34)* (3*10^8)/ (0.07*10^-9)
=2.8*10^-15 j
Then,
eV= E -----> V=E/e = (2.8*10^-15)/ (1.6*10^-19)
=17.5 kV.

But I don't think my approach is correct, so I am waiting your help.

2. Apr 4, 2017

### kuruman

Can you come up with abetter estimate? λm is to the left of the 0.04 nm mark.
What makes you say this?

3. Apr 4, 2017

OK, but it is λm of the bremsstrahlung radiation not the characteristic, isn't it?
I say that my approach is wrong because I don't actually understand the problem or how to start.

4. Apr 4, 2017

### kuruman

Y
Correct. Now suppose you have an electron accelerated to energy E. Consider two cases after it slams into the target: (1) it produces a single photon of energy E; (2) it produces two photons each with energy E/2. In which case is the wavelength shorter?

5. Apr 4, 2017

The photon in case (1) will have a shorter wavelength because it has a larger frequency. But what is the relation of this with what we try to get?

6. Apr 4, 2017

### kuruman

Note that we are trying to get the potential difference (1) through which the electron is accelerated.
Note that (1) is related to the energy of the electron (2).
Note that, if (2) goes into producing a single photon, this photon has the minimum wavelength λm (3).
Note that you can read (3) off the plot.

7. Apr 4, 2017

OK. But let me say something here, and I hope you get my point of view
The accelerated electron through the PD, can produce either a bremsstrahlung spectrum or a characteristic spectrum, right? The bremsstrahlung has its own minimum wavelength, the characteristic spectrum has its own minimum wavelength too, right?
The minimum wavelength of the bremsstrahlung can be estimated to be 0.035nm whilst the minimum wavelength of the characteristic is (in my opinion) directly below the left peak of the characteristic line spectrum, e.g., 0.07nm.
Are those points correct?

8. Apr 4, 2017

### kuruman

It is more appropriate to think that the accelerated electron produces a certain number of photons, not a spectrum. The sum of the energies of these photons must equal the initial energy of the electron. The spectrum is generated by looking at the distribution of a whole lot of photons produced by a whole lot of electrons.

9. Apr 4, 2017

OK, you are right. But still I don't know the answer of my question.

10. Apr 4, 2017

### kuruman

Please read post #6. It has all the hints you need. Connect the dots.

11. Apr 4, 2017