What are X-rays? Electrons are produced and used in an X-ray tube - Leaving Cert Physics - Question 10 - 2017

Electrons are produced via thermionic emission at the cathode or filament of the X-ray tube, where heat causes the electrons to be emitted into the vacuum.

To calculate the energy of an X-ray photon, we use the formula:

$E = hf$

where:

• $h = 6.63 \times 10^{-34} \text{ J s}$ (Planck's constant)
• $f$ is the frequency, which can be found using the speed of light: $f = \frac{c}{\lambda}$

Here, $abla = 0.02 \text{ nm} = 0.02 \times 10^{-9} \text{ m}$ and $c = 3 \times 10^8 \text{ m/s}$. Then, $f = \frac{3 \times 10^8 \text{ m/s}}{0.02 \times 10^{-9} \text{ m}} = 1.5 \times 10^{16} \text{ Hz}$

Substituting back into the energy equation: $E = (6.63 \times 10^{-34} \text{ J s})(1.5 \times 10^{16} \text{ Hz}) = 9.9 \times 10^{-15} \text{ J}$

The maximum velocity of an electron can be calculated using:

$E = \frac{1}{2} mv^2$

Rearranging for $v$ gives: $v = \sqrt{\frac{2E}{m}}$

Assuming an electron mass $m = 9.11 \times 10^{-31} \text{ kg}$ and using the energy calculated previously: $E = 9.9 \times 10^{-15} \text{ J}$

Thus, v = \sqrt{\frac{2 \cdot (9.9 \times 10^{-15})}{9.11 \times 10^{-31}}} = 1.48 \times 10^8 \text{ m/s}$$

The voltage can be determined using the relationship:

$V = \frac{E}{q}$

where $q = 1.6 \times 10^{-19} \text{ C}$ (the charge of an electron). Therefore:

$V = \frac{9.9 \times 10^{-15} \text{ J}}{1.6 \times 10^{-19} \text{ C}} = 62000 ext{ V}$

A labelled diagram of a photocell should include:

• Anode
• Semi-cylindrical cathode
• Case and vacuum

each labelled appropriately.

A photocell conducts current through the emission of electrons when exposed to light of suitable frequency. Light photons give sufficient energy to electrons in the cathode, allowing them to escape and create a flow of current towards the anode.

To find the number of electrons generated in the photocell:

• First, calculate the charge transferred in each minute using:

$Q = It$

where $I = 2 \mu A = 2 \times 10^{-6} A$ and $t = 60 s$: $Q = (2 \times 10^{-6} \text{ A})(60 s) = 1.2 \times 10^{-4} \text{ C}$

Next, to find the number of electrons: $n = \frac{Q}{q} = \frac{1.2 \times 10^{-4}}{1.6 \times 10^{-19}} = 7.5 \times 10^{14}$

Thus, approximately $7.5 \times 10^{14}$ electrons are generated in the photocell during each minute.