* (a) When in use, the neon gas between the electrodes emits electromagnetic radiation - Edexcel - A-Level Physics - Question 7 - 2023 - Paper 3

To determine which metal the light is incident upon, we need to calculate the energy of the emitted electrons using the equation:

$E = h f$

Where:

• $h = 6.63 \times 10^{-34} \text{J s}$
• $f = 5.56 \times 10^{14} \text{Hz}$

Calculating the energy:

$E = (6.63 \times 10^{-34}) (5.56 \times 10^{14}) = 3.69 \times 10^{-19} \text{J}$

Now, to find out which metal this energy corresponds to, we compare it with the work functions:

• Caesium: $3.36 \times 10^{-19} \text{J}$ (this is less than calculated energy)
• Potassium: $3.60 \times 10^{-19} \text{J}$ (this is also less than calculated energy)
• Sodium: $3.84 \times 10^{-19} \text{J}$ (this is greater than calculated energy)

The metal that the light is incident upon must be either Caesium or Potassium.