The first Nobel Prize in Physics was awarded in 1901 for the discovery of X-rays - Leaving Cert Physics - Question d - 2006

Electrons are emitted from the cathode by a process known as thermionic emission. This occurs when the cathode is heated, causing electrons to gain sufficient energy to break free from the metal surface.

Electrons are accelerated by the application of a high voltage between the anode and cathode. The potential difference provides the necessary electric field that acts on the electrons, increasing their kinetic energy as they move towards the anode.

The kinetic energy (K.E.) gained by an electron when it is accelerated through a potential difference (V) is given by the formula:

$E_k = qV$

where:

• charge of an electron, $q = 1.6 imes 10^{-19} C$
• potential difference, $V = 50 imes 10^3 V$

Substituting in the values:

$E_k = (1.6 imes 10^{-19} C)(50 imes 10^3 V)$

Calculating:

$E_k = 8.0 imes 10^{-15} J$

The minimum wavelength ($\lambda$) of the X-ray can be calculated using the equation:

$E = \frac{hc}{\lambda}$

where:

• $h = 6.6 \times 10^{-34} J s$ (Planck constant)
• $c = 3.0 \times 10^8 m s^{-1}$ (speed of light)
• $E = 8.0 \times 10^{-15} J$ (from the previous calculation)

Rearranging the formula to find $\lambda$ gives:

$\lambda = \frac{hc}{E}$

Substituting the values:

$\lambda = \frac{(6.6 \times 10^{-34} J s)(3.0 \times 10^8 m s^{-1})}{8.0 \times 10^{-15} J}$

Calculating:

$\lambda = 2.475 \times 10^{-11} m = 0.025 \text{ nm}$