Photoelectric Effect
A process of emitting the electrons from the a metal surface when the metal surface is exposed to an electro magnetic radiation of sufficiently high frequency is called Photoelectric effect.
For example: When ultraviolet light is given on the surface of an alkali metal, ejection of electrons takes place from surface of metal. This is the example of photo electric effect.
The electron ejected due to the photoelectric effect is termed as photoelectron and is denoted by e–.
The current produced as a result of the ejected photoelectrons is called photoelectric current.
The emission of number of photoelectrons and the kinetic energy of the ejected photoelectrons depend on the frequency of the light that is incident on the metal’s surface.
The photoelectric effect occurs when the electrons at the surface of the metal absorb energy from the incident light (radiation) and use it to overcome the attractive forces that bind them to the metallic nuclei.
Photoelectric cells
Metals that have low ionization enthalpies are used in Photoelectric cells.
The most popular metal which is used in Photoelectric cells is Cesium. which has the atomic number 55 and belongs to group 1 (alkali metals) and period 6.
Cesium is used widely in photocells as it can easily convert solar energy into electrical energy.
The work function is essentially the energy required to start the emission of electrons from the surface of a metal. It is represented by an ∅.
Photoelectric current depend :
The strength of the photoelectric current depends upon the intensity of the incident radiation. When the intensity of the radiation is more number of photons are striking on metal surface (more energy falls on metal surface ), so more electrons will be ejected i.e. photoelectric current will be increased.
The photoelectric effect can be explained by considering the particle nature of light
Planck found that the quantity of energy associated with each photon of particular frequency ν of light is E = h𝜈= hc/λ .It is is called Planck’s equation.
Where,
E denotes the energy of the photon
h is Planck’s constant
𝜈 denotes the frequency of the light
c is the speed of light (in a vacuum)
λ is the wavelength of the light
It is noted here that different frequencies of light carry photons of varying energies. For example, the frequency of photon of blue light is greater than that of red light (the wavelength of blue light is much shorter than the wavelength of red light).
Threshold Energy for the Photoelectric Effect
In order to occur photoelectric effect, the photons that incident on the surface of the metal must carry sufficient energy to overcome the attractive forces that bind the electrons to the nuclei of the metal. This amount minimum energy of incident photon is called Threshold energy (denoted by the symbol Φ) or it is called work function of metal.
The minimum frequency of light (photon)required for the photoelectric effect to occur is called threshold frequency.is usually denoted by the symbol 𝜈th, and the associated wavelength (called the threshold wavelength) is denoted by the symbol λth.
Work function =Φ = h𝜈th = hc/λth
How to calculate kinetic Energy of the emitted photoelectron.
Energy of photon = work function+ KE of photoelectron
hν = hν0 + ½ mv2
Where h = Plank’s constant (6.626 × 10^-34 J.s)
Frequently Asked Questions
Q: What does photoelectric current depend on?
Ans : The photoelectric current depends upon the intensity of the incident radiation.Higher the intensity of incident radiation more number of photon strike the metal surface and release more photoelectrons .Q: What is the term used for the minimum energy required to remove an photoelectron ?
Ans : Work function .Q: What is the time lag between the incidence of photons and the ejection of photoelectrons?
Ans : No time leg (approximetly less 1 ns) .Q: How does the intensity affect the photoelectric current?
Ans : As intensity increases, the photoelectric effect increases.Q: How does the intensity affect the photoelectric current?
Ans : As intensity increases, the photoelectric effect increases.Q: Light of wavelength 3500 Å is incident on two metals A and B. Which metal will yield more photoelectrons if their work functions are 5.5 eV and 2.5 eV respectively?
Ans : Metal B.Explanation :
Given, λ = 3500 Å = 3.5 X 10^-7m
Energy of incident photons = hv = hc/λ
= 3.536 eV
Work function of metal B is less than 3.536 eV.
