10.5.1 Production of X-rays

Production of X-rays

When the fast-moving electrons collide with the target, X-rays are produced by one of two methods
- Method 1: Bremsstrahlung
- Method 2: Characteristic Radiation

Method 1: Bremsstrahlung

When high-speed electrons collide with a metal target (often tungsten), they undergo a steep deceleration
- When a charged particle decelerates quickly, some of the energy released is converted into a photon
A small amount of the kinetic energy (~ 1%) from the incoming electrons is converted into X-rays as the electrons decelerate in the tungsten, due to conservation of energy
- The rest of the energy heats up the anode, which usually requires some form of cooling
The energy of the X-ray photon can be of any value, up to the original kinetic energy of the electron, giving a range of possible X-ray energies
- These X-rays cause the continuous or ‘smooth hump shaped’ line on an intensity wavelength graph

Ranges of Wavelengths in Bremsstrahlung Radiation

6-11-1-bremsstrahlung-graph_ocr-al-physics

The continuous spectra of Bremsstrahlung radiation at different acceleration potentials. As wavelength decreases, the energy of the X-rays photons increases.

When an electron is accelerated, it gains energy equal to the product of its charge and the accelerating potential, V , this energy can be calculated using:

E_max = eV

This is the maximum energy that an X-ray photon can have
The smallest possible wavelength is equivalent to the highest possible frequency and therefore, the highest possible energy
- This is assuming all of the electron’s kinetic energy has turned into electromagnetic energy

Therefore, the maximum X-ray frequency f_max, or the minimum wavelength λ_min, that can be produced is calculated using the equation:

$E_{m a x} = e V = h f_{m a x} = \frac{h c}{λ_{m i n}}$

The maximum X-ray frequency, f_max, is therefore equal to:

$f_{m a x} = \frac{e V}{h}$

The minimum X-ray wavelength, λ_min, is therefore equal to:

$λ_{m i n} = \frac{h c}{e V}$

Where:
- e = elementary charge (C)
- V = potential difference between the anode and cathode (V)
- h = Planck's constant (J s)
- c = the speed of light (m s⁻¹)

Method 2: Characteristic Radiation

Some of the incoming fast electrons cause inner shell electrons of the tungsten to be ‘knocked out’ of the atom, leaving a vacancy
- This vacancy is filled by an outer electron moving down and releasing an X-ray photon as it does (equal in energy to the difference between the two energy levels)
- Because these X-rays are caused by energy level transitions, they have only specific discrete energies
- They cause sharp spikes on an intensity wavelength graph
- The number of spikes depends on the element used for the target - there are two sets of spikes for a tungsten target, representing two sets of possible energy transitions

Characteristic Discrete X-Ray Wavelengths

6-11-1-characteristic-xray-graph_ocr-al-physics

Electron transitions emit photons with discrete energies. An incoming electron can cause these transitions, making tungsten emit characteristic photons.

Combined X-Ray Spectra of Bremsstrahlung and Characteristic Photons

6-11-1-xray-combined-graph_ocr-al-physics

Worked example

X-rays are a type of electromagnetic wave with wavelengths in the range 10⁻⁸ to 10⁻¹³ m

If the accelerating potential difference in an X-ray tube is 60 kV, determine if the photons emitted fall within this range.

Answer:

Step 1: Write out known quantities

Charge on an electron, e = 1.6 × 10⁻¹⁹ C
Accelerating potential difference, V = 60 000 V
Planck’s constant, h = 6.63 × 10⁻³⁴ J s
Speed of light, c = 3 × 10⁸ m s⁻¹

Step 2: Determine the maximum possible energy of a photon

The maximum possible energy of a photon corresponds to the maximum energy an electron could have:

E_max = eV

Step 3: Determine an expression for minimum wavelength

Planck relation: E = hf

Wave equation: c = fλ

When energy is a maximum:

E_max = eV = hf_max

Maximum energy corresponds to a minimum wavelength:

$e V = \frac{h c}{λ_{m i n}}$

Rearrange for minimum wavelength, λ_min:

$λ_{m i n} = \frac{h c}{e V}$

Step 4: Calculate the minimum wavelength λ_min

$λ_{m i n} = \frac{(6.63 \times 10^{- 34}) (3 \times 10^{8})}{(1.6 \times 10^{- 19}) (60 000)}$

λ_min = 2.1 × 10⁻¹¹m

Step 5: Comment on whether this is within the range for the wavelength of an X-ray

X-ray wavelengths are within 10⁻⁸ to 10⁻¹³ m
The minimum wavelength for a 60 kV supply is 2.1 × 10⁻¹¹m, which means the photons produced will be X-rays

Worked example

A typical spectrum of the X-ray radiation produced by electron bombardment of a metal target is shown below.

Explain why:

(a)

A continuous spectrum of wavelengths is produced.

(b)

The gradient is steeper at shorter wavelengths.

(c)

The spectrum has a sharp cut-off at short wavelengths.

Answer:

Part (a)

Step 1: Consider the path of the electrons from the cathode to the anode

Photons are produced whenever a charged particle undergoes a large acceleration or deceleration
X-ray tubes fire high-speed electrons at a metal target
When an electron collides with the metal target, it loses energy in the form of an X-ray photon as it decelerates

Step 2: Consider the relationship between the energy of the electron and the wavelength of the photon

The wavelength of a photon depends on the energy transferred by a decelerating electron
The electrons don't all undergo the same deceleration when they strike the target
This leads to a distribution of energies, hence, a range, or continuous spectrum, of wavelengths is observed

Part (b)

Step 1: Identify the significance of the intensity

The intensity of the graph signifies the proportion of photons produced with a specific energy, or wavelength
The higher the intensity, the more photons of a particular wavelength are produced
In other words, the total intensity is the sum of all the photons with a particular wavelength

Step 2: Explain the shape of the graph

When a single electron collides with the metal target, a single photon is produced
Most electrons only give up part of their energy, and hence there are more X-rays produced at wavelengths higher than the minimum (or energies lower than the maximum)
At short wavelengths, there is a steeper gradient because only a few electrons transfer all, or most of, their energy

Part (c)

Step 1: Identify the relationship between minimum wavelength and maximum energy

The minimum wavelength of an X-ray is equal to

$λ_{m i n} = \frac{h c}{E_{m a x}}$

The equation shows the maximum energy of the electron corresponds to the minimum wavelength, they are inversely proportional

$λ_{m i n} \propto \frac{1}{E_{m a x}}$

Therefore, the higher the energy of the electron, the shorter the wavelength of the X-ray produced

Step 2: Explain the presence of the cut-off point

The accelerating voltage determines the kinetic energy which the electrons have before striking the target
The value of this accelerating voltage, therefore, determines the value of the maximum energy
This corresponds to the minimum, or cut-off, wavelength

AQA A Level Physics

Revision Notes

Production of X-rays

Method 1: Bremsstrahlung

Ranges of Wavelengths in Bremsstrahlung Radiation

Method 2: Characteristic Radiation

Characteristic Discrete X-Ray Wavelengths

Combined X-Ray Spectra of Bremsstrahlung and Characteristic Photons

Worked example

Worked example

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Author: Katie M