Subsection 7.1.2 Photoelectric Effect
The photoelectric effect is a phenomenon in which electrons are emitted from a metal surface when light of a certain frequency shines on it. Electrons are ejected because of the transfer of energy from light photons to electrons in the metal. The photoelectric effect demonstrated that light has a dual nature, both as a wave and as a particle.
The photoelectric effect was first observed by Hertz in the late 19th century and later explained by Einstein in 1905. The basic idea behind the photoelectric effect is that light is composed of individual particles, called photons, that carry a specific amount of energy. When these photons hit a metal surface, they transfer their energy to the electrons in the metal. If the energy of the photon is high enough, it can cause an electron to be ejected from the metal and become a free electron. This process only occurs with light of a certain frequency, known as the
threshold frequency. If the frequency of the light is below this threshold, no electrons will be emitted, no matter how intense the light is. If the frequency is above the threshold, the number of electrons emitted will increase with the intensity of the light. The photoelectric effect challenged the classical understanding of light as a wave, and led to the development of quantum mechanics, which views light and matter as having both wave-like and particle-like properties. The photoelectric effect is widely used in photodiodes, photovoltaics, and photoelectric sensors, etc.
Photoelectric Effect Simulation.
Einstein’s formula for photoelectric effect is given by
\begin{equation}
hf = KE + w\tag{7.1.2}
\end{equation}
where \(hf\) is energy of a photon of light whose frequency is \(f,\) \(KE\) is the kinetic energy of the emitted electron, and \(w\) is work function of the metal (i.e., energy needed to pull the electron from the metal). Eventhough,photon is massless and always moves with the speed of light, it behaves like a particle and possesses energy and momentum. It also interacts with other particles in the same way as one particle interacts with others.
Example 7.1.4.
The ultraviolet light of frequency \(3. 0 \times 10^{15} Hz\) is incident on the surface of a metal, whose work function is \(1.8 eV\text{.}\) Show that the kinetic energy of the emitted electrons is \(1.7\times 10^{-18} J\text{.}\)
Solution.
Given: \(f=3. 0 \times 10^{15} \,Hz\text{,}\) \(h=6.63\times 10^{-34} (\,Js)\text{,}\) \(w=1.8\,eV= 1.8\times 1.67\times 10^{-19}C= 3.0\times10^{-19}J\text{.}\)
\begin{align*}
E=hf \amp = KE+w\\
or, \quad KE \amp = hf-w\\
or, \quad KE \amp = 6.63\times 10^{-34} \times 3. 0 \times 10^{15} -3.0\times10^{-19}\\
\therefore KE \amp = 19.89\times 10^{-19} -3.0\times10^{-19}= 16.89\times10^{-19} \approx 1.7\times10^{-18} J
\end{align*}
