The explanation of the photoelectric effect by Albert Einstein led to the quantum revolution in physics - Leaving Cert Physics - Question c - 2022

To calculate the maximum speed of the emitted electrons, we use the equation derived from Einstein's photoelectric equation:

$E_k = hf - ext{Work Function}$

Where:

• $h$ is Planck's constant ($6.63 \times 10^{-34} \, J \, s$)
• $f$ is the frequency of the light
• The work function ( \phi ) is given as 2.4 eV, which can be converted to Joules: ( \phi = 2.4 \times 1.6 \times 10^{-19} , J = 3.84 \times 10^{-19} , J )

First, convert wavelength to frequency: [ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{450 \times 10^{-9}} = 6.67 \times 10^{14} , Hz ]

Now, substituting into the kinetic energy equation: [ E_k = (6.63 \times 10^{-34})(6.67 \times 10^{14}) - 3.84 \times 10^{-19} ] [ E_k = 4.42 \times 10^{-19} , J ]

The maximum speed ( v ) can be found using: [ E_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}} ] Here, the mass of the electron, ( m = 9.11 \times 10^{-31} , kg ) [ v = \sqrt{\frac{2(4.42 \times 10^{-19})}{9.11 \times 10^{-31}}} = 3.56 \times 10^6 , m/s ]

Incident Energy Decreases

As the wavelength of the incident light increases, the energy of the photons decreases since the energy is inversely proportional to the wavelength, expressed by the formula: [ E = \frac{hc}{\lambda} ]

When the energy of the incident light falls below the work function of the metal, electrons cannot be emitted.

Threshold Frequency

There is a threshold frequency associated with the work function of the metal. When the frequency of the incident light falls below this threshold frequency, no electrons are emitted regardless of the intensity of the light. This explains why increasing the wavelength to a certain point results in no electron emission.