What is thermionic emission? X-rays are produced when high-energy electrons collide with a target - Leaving Cert Physics - Question 9 - 2010

To draw a labelled diagram of an X-ray tube, include the following components:

• Vacuum: Indicates the absence of air to prevent electron scattering.
• Cathode: The negative electrode that emits electrons when heated.
• Anode/Target: The positive electrode where electrons collide and produce X-rays.
• High Voltage Supply: Indicates the voltage that accelerates electrons towards the anode.
• Shielding and Cooling: Surrounding structures to protect against radiation and manage heat.

X-rays are a form of electromagnetic radiation that consists of photons or quanta of energy. They have a high frequency and a short wavelength compared to visible light. The significant differences are:

• Electromagnetic Radiation: X-rays carry more energy than visible light.
• High Frequency/Short Wavelength: Their wavelengths are typically in the range of 0.01 to 10 nanometers, making them much shorter than those of light rays.
• Penetration Ability: X-rays can penetrate materials that visible light cannot, a property that is useful in medical imaging and security.

1. Medical Imaging: X-rays are extensively used in radiography to diagnose bone fractures and other medical conditions.
2. Security Scanning: X-rays are employed in airports for imaging luggage to check for prohibited items.

In an X-ray tube, when high-energy electrons hit the target, only a small percentage of their energy is converted into X-rays. The majority of the energy is converted into heat. This means that the target material must have a high melting point to withstand the heat generated during the process, typically requiring materials such as tungsten which has a melting point of over 3400°C.

To calculate the maximum energy (W) of an electron as it hits the target, we can use the formula:

$W = qV$

where:

• $q$ is the charge of the electron ($1.6 \times 10^{-19} C$)
• $V$ is the potential difference (40 kV = $40 \times 10^{3} V$)

Calculating gives: $W = (1.6 \times 10^{-19})(40 \times 10^{3}) = 6.4 \times 10^{-15} J$

The frequency (f) of the most energetic X-ray can be calculated using the equation:

$E = hf$

where:

• $E$ is the energy of the X-ray ($6.4 \times 10^{-15} J$)
• $h$ is Planck's constant ($6.6 \times 10^{-34} J s$)

Rearranging gives: $f = \frac{E}{h}$

Plugging in the values: $f = \frac{6.4 \times 10^{-15}}{6.6 \times 10^{-34}} = 9.7 \times 10^{18} Hz$