In a demonstration of the photoelectric effect, electromagnetic radiation of frequency $f$ was incident on the surface of a metal - Edexcel - A-Level Physics - Question 11 - 2023 - Paper 2

When a photon with frequency $10.0 imes 10^{14} Hz$ strikes the metal surface, it causes a photoelectron to be released. The energy of the photon can be calculated using the equation:

$E = hf$

Substituting the values: $E = (6.626 imes 10^{-34} J ext{s}) (10.0 imes 10^{14} Hz) = 6.626 imes 10^{-20} J$

To find the maximum kinetic energy ($E_{max}$) of the emitted photoelectron, we can write: $E_{max} = E - ext{Work function}$

Assuming the work function for the metal is available from the table, we would subtract the work function from the photon energy. However, to calculate the maximum velocity we use: E_{max} = rac{1}{2} mv_{max}^2

Solving for $v_{max}$, v_{max} = ext{sqrt}igg(rac{2E_{max}}{m}igg)

We need the mass of the electron, $m = 9.11 imes 10^{-31} kg$.

Once the $E_{max}$ is determined, substitute it to find $v_{max}$.

To determine which metal was used in the demonstration, we can compare the calculated maximum kinetic energy (derived from the photon energy minus the work function) with the provided work functions for the three metals. The work function values are as follows:

• Caesium: $φ = 2.2 ext{ eV}$
• Zinc: $φ = 4.3 ext{ eV}$
• Beryllium: $φ = 5.0 ext{ eV}$

Given the photon energy calculated previously, if the energy falls above the work function of one of these metals but below the others, we can deduce that the metal used in the demonstration has the corresponding work function that matches the emitted photoelectrons' energy.