A simplified diagram of an apparatus for an experiment to investigate the photoelectric effect is shown below - NSC Physical Sciences - Question 11 - 2017 - Paper 1

The reading on the ammeter will INCREASE. This is because an increase in intensity means that the number of photons incident per unit time increases. Consequently, more electrons will be emitted per unit time from the cathode of the photoelectric tube, leading to a higher photocurrent.

To find the maximum wavelength (threshold wavelength), we use the equation relating energy and frequency:

$E = hf$

Where:

• $E$ is the energy of the photon,
• $h$ is Planck's constant ($6.63 \times 10^{-34} \, \text{Js}$),
• $f$ is the frequency of the incident light.

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

Substituting the values: $\lambda = \frac{6.63 \times 10^{-34} \, \text{Js}}{5.9 \times 10^{14} \, \text{Hz}} = 1.12 \times 10^{-19} \text{m}$

The photoelectric equation is given by:

$E = W + E_{k}$

Where:

• $E$ is the energy of the incident photon,
• $W$ is the work function of the material,
• $E_k$ is the kinetic energy of the emitted electron.

This equation shows that as the energy of the light increases (by increasing the frequency or reducing the wavelength), more energy is available for the electrons, increasing their kinetic energy. Therefore, light with higher energy results in more energetic photoelectrons, establishing a direct relationship between the frequency of light and the kinetic energy of emitted electrons.