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What's the difference between the effect caused by the number of photons of an x ray beam increasing and the effect caused by the individual photon energies of an x ray beam increasing?
Don't they both make an x ray beam more powerful?
This question propped up in my mind whilst I was reading this-
Source I used is below:
Composition of an X-ray device:
X-ray instruments are made of electrical parts and mechanical parts:
The X-ray tube is fed with high voltage. This high voltage comes from a high voltage transformer and a rectifier. The transformer converts normal mains voltage into a high and continuously adjustable voltage for the X-ray tube. Electrons need to pass through the X-ray tube in one direction, therefore the rectifier converts any AC to DC. However, the rectifier provides a pulsating current that needs to be offset by ‘smoothing’ (otherwise, undesired low energy photons might be produced).
The X-ray tube is where the X-rays are produced once provided with high voltage. The X-ray tube contains a hot cathode (filament) and a cold anode and electrons are accelerated between them, getting slowed down by tungsten at the anode. This releases kinetic energy and a small amount of that energy turns into high energy photons (the X-rays!). The X-rays are emitted from an area on the anode called the focus. The production of photons like this from the deceleration of charged particles is known as Bremsstrahlung radiation (this is deceleration). The energy of a decelerated electron at the anode (U × e) is given by:
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U×e = energy of incident electron at anode
h = Planck’s constant (6.62 x 10 -34)
fmax = maximum frequency
c = propagation speed
λmin = smallest wavelength
It is worth noting that the current from the cathode to the anode in the X-ray tube can be adjusted. This is done by increasing the heating of the cathode – this causes the photon flow rate of the X-ray beam produced to increase, but individual photon energies do not increase. To increase individual photon energies (making the X-ray beam more powerful), the voltage sent to X-ray tube from the transformer has to be increased.
Don't they both make an x ray beam more powerful?
This question propped up in my mind whilst I was reading this-
Source I used is below:
Composition of an X-ray device:
X-ray instruments are made of electrical parts and mechanical parts:
- o Electrical parts include a source of high voltage [transformer + rectifier], an X-ray tube [containing electrodes] and a control panel. The control panel is located outside the room behind a shield of glass and lead walls to protect the physician/assistant from the ionising radiation.
- o Mechanical parts include a stand for the X-ray tube, an examination table for the patient, a cassette (containing the radiographic film to record the X-ray) and a Bucky grid (aka collimator) under the patient (which removes scattered electrons), and maybe an image intensifier (for fluorescent recording of X-rays)
The X-ray tube is fed with high voltage. This high voltage comes from a high voltage transformer and a rectifier. The transformer converts normal mains voltage into a high and continuously adjustable voltage for the X-ray tube. Electrons need to pass through the X-ray tube in one direction, therefore the rectifier converts any AC to DC. However, the rectifier provides a pulsating current that needs to be offset by ‘smoothing’ (otherwise, undesired low energy photons might be produced).
The X-ray tube is where the X-rays are produced once provided with high voltage. The X-ray tube contains a hot cathode (filament) and a cold anode and electrons are accelerated between them, getting slowed down by tungsten at the anode. This releases kinetic energy and a small amount of that energy turns into high energy photons (the X-rays!). The X-rays are emitted from an area on the anode called the focus. The production of photons like this from the deceleration of charged particles is known as Bremsstrahlung radiation (this is deceleration). The energy of a decelerated electron at the anode (U × e) is given by:
U×e = energy of incident electron at anode
h = Planck’s constant (6.62 x 10 -34)
fmax = maximum frequency
c = propagation speed
λmin = smallest wavelength
It is worth noting that the current from the cathode to the anode in the X-ray tube can be adjusted. This is done by increasing the heating of the cathode – this causes the photon flow rate of the X-ray beam produced to increase, but individual photon energies do not increase. To increase individual photon energies (making the X-ray beam more powerful), the voltage sent to X-ray tube from the transformer has to be increased.
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#2
A simplistic answer (sorry!)
Increasing beam current -> more photons (of same energy) -> 'brighter' image (like taking a photo on a sunny day)
increasing voltage -> photons of higher energy -> photons of shorter wavelength -> more penetrating.
Experts may be able to give accurate explanations!
Increasing beam current -> more photons (of same energy) -> 'brighter' image (like taking a photo on a sunny day)
increasing voltage -> photons of higher energy -> photons of shorter wavelength -> more penetrating.
Experts may be able to give accurate explanations!
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(Original post by old_teach)
A simplistic answer (sorry!)
Increasing beam current -> more photons (of same energy) -> 'brighter' image (like taking a photo on a sunny day)
increasing voltage -> photons of higher energy -> photons of shorter wavelength -> more penetrating.
Experts may be able to give accurate explanations!
A simplistic answer (sorry!)
Increasing beam current -> more photons (of same energy) -> 'brighter' image (like taking a photo on a sunny day)
increasing voltage -> photons of higher energy -> photons of shorter wavelength -> more penetrating.
Experts may be able to give accurate explanations!
By increasing the heating of the cathode, then I assume the number of electrons being released by the cathode increases or in other words the intensity of the electric current (or the intensity of the electrons) increases. But wait, doesn't oxidation happen at the anode?!
Your amazingly put simplistic answer was very helpful. Can you let me know about the above if possible please?
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#4
What you've said is correct, I think.
It's all in a vacuum, so no chance of oxidation!
It's all in a vacuum, so no chance of oxidation!
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(Original post by old_teach)
What you've said is correct, I think.
It's all in a vacuum, so no chance of oxidation!
What you've said is correct, I think.
It's all in a vacuum, so no chance of oxidation!
Thank you!
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#6
Yes - the cathode is heated so it emits electrons and they (electrons) are accelerated by 1000s of volts.
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(Original post by old_teach)
Yes - the cathode is heated so it emits electrons and they (electrons) are accelerated by 1000s of volts.
Yes - the cathode is heated so it emits electrons and they (electrons) are accelerated by 1000s of volts.
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