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X-rays
X-ray Tube
K = Hot filament cathode
A = Tungsten anode
Uh = Heater Voltage (e.g. 12V)
Ua = Accelerating voltage (e.g. 50kV)
W = Cooling water
X = X-rays
Clear glass tube containing vacuum
Q. Explain the following...
i. the source of high energy electrons that bombard the tungsten anode.
ii. the two mechanisms by which X-rays are created in the tube.
iii. how it is possible to increase the intensity of the emitted X-rays
iv. how it is possible to decrease the minimum wavelength of the emitted X-ray photons (hint: this involves increasing the maximum frequency)
i. Electrons are emitted by thermoelectric emission from the hot cathode. They are then accelerated through a large p.d. (50kV) and collide with the tungsten anode.
ii. X rays are emitted in two ways...
The decelerating electrons may emit X-rays. The decelerating electrons may also excite electrons in the tungsten anode. When these fall back to lower energy levels they also emit X-rays (or other em radiation).
iii. Increasing cathode temperature increases thermoelectric emission and thus increases X-ray photon intensity
iii. Increasing the accelerating voltage increases electron energy and thus increases max X-ray photon energy and decreases min wavelength.
Features of the X-ray spectrum
Continuous spectrum (Bremsstrahlung):
This is due to the decelerating electrons. Different electrons have different amounts of energy and thus different wavelength photons are emitted.
Characteristic Peaks:
These are due to atomic electrons falling back to the lowest energy level after being excited by a collision from an accelerated electron. Thus they are characteristic of the element of which the target is constructed (e.g. tungsten).
Minimum Wavelength of X-rays
Energy of accelerated electron = eV
Energy of X-ray photon = hf
Thus if all the energy of the electron is transferred to the photon then we can say...
So the greater the accelerating p.d. the smaller the minimum wavelength of the emitted radiation.
Q. Sketch two X-ray spectra for the same target element but showing two different accelerating p.d.’s.
eV = hf or eV = hcλ