Electromagnetic Waves

G5 - X Rays

Coolidge tube (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 number of electrons produced thus increasing number of X-ray photons produced

iii. Increasing the accelerating voltage increases electron energy and thus increases max X-ray photon energy and decreases min wavelength.

X-ray spectrum

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.

Features of the X-ray spectrum

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.

eV = hf or eV = hcλ

Q. Sketch two X-ray spectra for the same target element but showing two different accelerating

Q. Sketch two X-ray spectra for the same target element but showing two different accelerating

• What is the minimum wavelength of X-rays emitted in a 50 kV tube?

What is the minimum wavelength of X-rays emitted in a 50 kV tube?

hf = eV

f = eV/h = (1.6 x 10-19 x 50000)/6.63 x 10-34

f = 1.21 x 1019 Hz

λ = c/f = 3 x 108/1.21 x 1019 = 2.49 x 10-11 m

X – ray diffraction/Bragg reflectionThe diagram below shows X-rays being reflected from a crystal. Each layer of atoms acts like a mirror and reflects X-rays strongly at an angle of reflection that equals the angle of incidence. The diagram shows reflection from successive layers

 

If the path difference between the beams from successive layers of atoms is a whole number of wavelengths, then there is constructive interference. Path difference is the distance AB + BC as above:

(AB + BC) = 2dsinθ. The reflected beam has maximum intensity when 2dsinθ = nλ

• X-rays of wavelength 1.45 x 10-10 m are reflected off a crystal. When the angle between the X-ray beam and the face of the crystal is increased from zero, a strong reflected beam is detected when the angle becomes 36.4º.

• X-rays of wavelength 1.45 x 10-10 m are reflected off a crystal. When the angle between the X-ray beam and the face of the crystal is increased from zero, a strong reflected beam is detected when the angle becomes

36.4º. Calculate the spacing of the crystal planes responsible for this reflection.

• X-rays of wavelength 1.45 x 10-10 m are reflected off a crystal. When the angle between the X-ray beam and the face of the crystal is increased from zero, a strong reflected beam is detected when the angle becomes

36.4º. Calculate the spacing of the crystal planes responsible for this reflection.

• Using the Bragg condition we find d = nλ/2sinθ• d = (1x1.45 x 10-10)/2sin36.4º = 1.22 x 10-10 m

• X-rays of wavelength 1.45 x 10-10 m are reflected off a crystal. When the angle between the X-ray beam and the face of the crystal is increased from zero, a strong reflected beam is detected when the angle becomes

36.4º. Calculate the spacing of the crystal planes responsible for this reflection. d = = 1.22 x 10-10 m

• Are there any other angles at which strong reflection is observed?

• X-rays of wavelength 1.45 x 10-10 m are reflected off a crystal. When the angle between the X-ray beam and the face of the crystal is increased from zero, a strong reflected beam is detected when the angle becomes

36.4º. Calculate the spacing of the crystal planes responsible for this reflection. d = = 1.22 x 10-10 m

• Are there any other angles at which strong reflection is observed?

• With the value of d we have just calculated,• sinθ = nλ/2d = n x 0.59• No other angles exist, since for n = 2, sinθ >1, which is

impossible

Structure of crystals

• By measuring the spacing of different planes in a crystal it is possible to deduce the structure of the crystal

Crystal structures

DNA structure

Questions

• Page 638 Questions 2, 4, 7, 9.