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einstein photoelectric equation class 12 notes explain , write down explain the concept of threshold frequency

By   March 29, 2023

write down einstein’s photoelectric equation and then explain the concept of threshold frequency einstein photoelectric equation class 12 notes explain ?

Photoelectric Effect

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When electromagnetic radiation of appropriate frequency falls on a metal, electrons are emitted. This phenomenon is called photoelectric effect and the emitted electrons are called photoelectrons because they are liberated by means of light.

Einstein’s Photoelectric Equation

The classical electromagnetic wave theory of light, which successfully explained interference, diffraction, and polarization of light, could not account for the observations related to photoelectric effect. In 1900 Planck postulated that light waves consist of tiny bundles of energy called photons or quanta. The energy of a light wave of frequency n is given by E = hv, where h is Planck’s constant. Photon is simply a light wave of energy E. Following Planck’s idea, Einstein proposed a theory for photoelectric effect. According to him, when a photon of light falls on a metal, it is absorbed, resulting in the emission of a photoelectron. The maximum kinetic energy

This is the famous Einstein’s photoelectric equation. The term hn represents the total energy of the photon incident on the metal surface. The photon penetrates a distance of about 10–8 m before it is completely absorbed. In disappearing, the photon imparts all its energy to a single electron. Part of this energy is used up by the electron in freeing itself from the atoms of the metal. This energy designated by W0 in Eq. (1) is called the work-function of the metal and is a characteristic of it. The rest of the energy is used up in giving the electron kinetic energy.

The work function W0, i.e. the energy required to pull an electron away from the surface of the metal, is large for heavier elements like platinum whereas for other elements like alkali metals, W0 is quite small. The minimum, or threshold energy which a photon must have to free the electron from the surface of the metal should be equal to its work function. If the threshold frequency is n0, the threshold energy is hv0. Thus

It is evident that when v < v0, no electron is emitted for any intensity of light. When v > v0, the energy of the electron increases linearly with the frequency v of light. Since intensity of light is a measure of the number of photons and since each photon emits a photoelectron on absorption, the intensity of photoelectrons is proportional to the intensity of light. Below a certain negative voltage v0, no photoelectrons are emitted no matter what the intensity of light is. This voltage is called the cut-off or stopping potential. Since there is no photoelectric emission at potentials less than v0, the maximum velocity vmax acquired by the photoelectrons is given by

Laws and graphs of photoelectric effect

(1) For a given emitter illuminated by radiation of a given frequency, the photoelectric current is proportional to the intensity of radiation

(2) The maximum kinetic energy (Kmax) of photoelectrons is proportional to the frequency (v) of the
incident radiation and is independent of intensity of
the radiation

Kmax = h(v – v0)
Slope of graph = h(Planck’s constant).
Kmax = 0 when v ≤ v (threshold frequency)

(3) For every emitter there is definite threshold frequency (v0) below which no photoelectrons are emitter no matter what the intensity of radiation is.

(4) Graph of stopping potential (v0) versus frequency n of incident radiation

Slope of graph = h/e , which is the same for all metals.

(5) Graph of photoelectric current (i) versus voltage (V) for radiations of different intensities (I1 > I2) but of the same frequency.

(6) Graph of photoelectric current (i) versus voltage (V) for radiations of different frequencies (v1 > v2) but of the same intensity,

(7) Threshold wavelength is λ0 = c/vFor photoelectric emission λ < λ0.

Example 3 The maximum wavelength of radiation that can cause photoelectric emission in a metal of work function 2.5 eV is very nearly equal to

(a) 400 nm (b) 500 nm

(c) 600 nm (d) 700 nm

answer  = 497 nm So the correct choice is (b).

2. Wave Nature of Matter

In 1924, Louis de Broglie, a French theoretical physicist, derived an equation which predicted that all atomic particles have associated with them waves of a definite wavelength. Under certain circumstances, a beam of electrons or atoms will behave like a group of waves. On the basis of theoretical considerations, de Broglie predicted that the wavelength λ of these waves is given by

λ = h/p = h/mv

where h is Planck’s constant and p is the momentum of the particles. This equation is known as de Broglie’s wave equation. For an electron moving at a high speed, the momentum is large and the wavelength λ is small. The faster the electron, the shorter is the wavelength. Notice that the particle need not have a charge to have an associated wave. This is why de Broglie waves are sometimes referred to as matter waves

1. If the rest mass of a particle is m0, its de Broglie wavelength is given by

2. In terms of kinetic energy K, de Broglie wavelength is given by

3. If a particle of charge q is accelerated through a potential difference V, its de Broglie wavelength is given by

4. For a gas molecule of mass m at temperature T kelvin, the de Broglie wavelength is given by

where k is the Boltzmann constant.