# Need help with X-ray production

• Leaping antalope
In summary, the graph is a representation of the intensity of x-rays generated from a potential difference collision between electrons and a target. The x-ray intensity peaks at short wavelengths and decreases as the wavelength increases. The potential difference is used to accelerate the electrons and the new graph is created when the potential difference is half of the original. The new graph is on a photon energy scale and is similar to the original but may have a missing characteristic radiation peak. The difference between A and C is that C is bremsstrahlung radiation while A is continuous radiation. The intensity of C is smaller than A. Lastly, when the potential is half of the original, the entire graph shifts to the right and the peaks and continuous radiation
Leaping antalope
Need help with "X-ray production"

Hi everyone. i am in 12th grade now.
im having problems with the production of X-ray...
Can someone tell me how X-ray is generated? From the very begining...to the charateristic raditaoin and the Bremsstrahlung Radiation...to the intensity graph (the one with two peaks).

I don't understand why the graph is like that...i mean, what is the relationship between wavelength and x-ray intensity?

I also don't get the calculation: when do we use qV=1/2 mv^2, and when do we use qV=hf? (q is the charge of electrons, V is the potential difference, 1/2mv^2 is the kinetic energy, h is plank's constant, f is frequency.)

And how can we find out the followings if only an x-ray intensity graph is given?
1. the potential difference
2. the difference of energy between the ground state and the first excited state?

And...if a rough x-ray graph is given...what will the new graph be when the potential difference is half?

I'm going to take the IB physics exam soon in May(similar to AP in US, A-level in UK) and i still don't get this X-ray stuff. Can somebody help me? thanks a lot!

Leaping antalope said:
I don't understand why the graph is like that...i mean, what is the relationship between wavelength and x-ray intensity?
The only thing you need to worry about is at what wavelengths the intensity is higher than zero (the cutoff at short wavelengths).
I also don't get the calculation: when do we use qV=1/2 mv^2, and when do we use qV=hf? (q is the charge of electrons, V is the potential difference, 1/2mv^2 is the kinetic energy, h is plank's constant, f is frequency.)
The kinetic energy of the electrons after they have been accelerated by the potential diffference V is qV. That is also the highest energy of the photons that can be produced. (Photon energy is hf.)[/QUOTE]

X-rays are generated by using a potential difference to accelerate and collide electrons into a target. This also answer's when you're supposed to use qV=1/2 mv^2. You use qV=hf to find out what's the maximum energy of a photon that can be produced when the electrons collide to the target.

Bremsstrahlung, the breaking radiation, is a product of the elctrons slowing down. It creates the continuous part of the spectrum. The characteristic radiation is a property of the target you collide the electrons into. The colliding electrons with suffcient energy can knock off electron's from one of the target atom species' inner shells causing an electron from one the outer shells to "fall in" and take it's place emitting an photon in the x-ray wavelength area.

1: Weird question. Usually folks now what accelerating voltage they're using. From the spectrum you should see a point where the intensity starts to differ from 0. This point corresponds to the maximum energy a photon generated by the collision. eV=h/lambda_min, solve for V.

2. These you can calculate from the positions of the characteristic peaks.

The new graph is of similar form but it may be missing the charcteristic peak if the electrons do not have enough energy to generate characteristic radiation via the collisions. And lambda_min is larger.

inha said:
2. These you can calculate from the positions of the characteristic peaks.

so...should i find out the wavelengths corresponding to the two peaks, and use E=hf1-hf2 (c=frequency x wavelength) ? is this the energy difference b/t the ground state and the first excited state?

Also...please see the graph attached...A is the continuous radiation, B is the charasteristic radiation...then how about C? is C also the continuous radiation? what's the difference between A and C?

And when the potential is half (suppose that half V still has enough energy to remove the inner-shell electrons and produce the charactereistic radiation...), will the entire graph shift to the right (wavelength_min larger)? will the peaks and the continuous radiation be of smaller intensity?

Thanks~

#### Attachments

• xray2.PNG
1.6 KB · Views: 512
Leaping antalope said:
what's the difference between A and C?
Both are bremsstrahlung.
And when the potential is half (suppose that half V still has enough energy to remove the inner-shell electrons and produce the charactereistic radiation...), will the entire graph shift to the right (wavelength_min larger)? will the peaks and the continuous radiation be of smaller intensity?
The graph is on a photon energy scale. It looks like a silver spectrum taken at 35 kV. At half the voltage, no holes can be created in the K-shell of silver.

Thanks...
oh, another question...for the bremsstrahlung radiation part, the projectile eletrons avoid all outer-shell electrons of the target and interact with the nucleus of the target. because the nucleus is of positive charge, the projectile electrons will be attracted. How come the electrons slow down? i thought they should be accelerating toward the nucleus due to the stronger attraction force from the protons.

In classical terms: when high-energy electrons get close to nuclei, they are deflected (think of a hyperbolic trajectory). This change of the direction of the electron is an acceleration (whatever the sign), and classically, an accelerated charge must radiate. So there will be a loss of kinetic energy.

But when the spectrum is studied at very high resolution, one can see at the highest photon energies the unoccupied part of the density of states of the anode material, and (if this is a metal) the Fermi level cutoff. This is studied in BIS (Bremsstrahlung Isochromat Spectroscopy). And that requires a quantum-mechanical explanation.

Last edited:
I see. Thanks Pieter Kuiper and Inha. I just bombed my physics mid-term today...i just hope that i can do better on the real IB exam.

Oh, and by the way...do anyone know any useful website about particle physics (leptons, mesons, bosons...and all those)...cuz this is also my weak point. Thanks~

Check out the sticky thread at the nuclear and particle physics forum here. IIRC it should go through the basics and it had some nice link to some particle physics page which doesn't require that much previous knowledge on the subject.

## What is X-ray production?

X-ray production is the process of generating X-rays, a type of electromagnetic radiation, for various purposes such as medical imaging, security screening, and industrial inspection.

## How are X-rays produced?

X-rays can be produced through two main methods: by accelerating charged particles such as electrons and ions using high-voltage electricity, or by bombarding a metal target with high-energy electrons in a process called bremsstrahlung.

## What are some applications of X-rays?

X-rays have many practical applications in different fields. In medicine, they are used for diagnostic imaging and cancer treatment. In industry, X-rays are used for non-destructive testing and quality control. In security, they are used for baggage and cargo screening. They are also used in research, astronomy, and material science.

## What are the risks associated with X-ray production?

X-rays are a form of ionizing radiation, which means they can cause damage to living cells. Exposure to high levels of X-rays can lead to radiation sickness and an increased risk of cancer. However, the risks of X-ray production can be minimized with proper safety measures and equipment.

## What are the advancements in X-ray production technology?

With advancements in technology, X-ray production has become more efficient and safer. New techniques such as digital radiography and computed tomography have improved image quality and reduced radiation exposure. Portable and handheld X-ray devices have also been developed for easier and faster use in various settings.

• Optics
Replies
54
Views
5K
• Optics
Replies
6
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
7
Views
367
• High Energy, Nuclear, Particle Physics
Replies
17
Views
2K
• Optics
Replies
1
Views
1K
• Optics
Replies
3
Views
2K
• Atomic and Condensed Matter
Replies
1
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
6
Views
1K
• Other Physics Topics
Replies
3
Views
3K
• High Energy, Nuclear, Particle Physics
Replies
11
Views
1K