State laws of photoelectric effect and derive the relation h v = W + 1 2 m v m a x 2 where W is work function, h is Planck constant, m is mass of electron, v m a x is velocity.

Einstein explained photoelectric effect on the basis of quantum theory. The main points are 1. Light is propagated in the form of bundles of energy.called quanta having energy given by the Planck's relation. E = hν Where ν = frequency and h = Planck's constant 2 The photoelectric effect is due to collision of a photon of incident light and a bound electron of the metallic cathode. The interaction between the incident quanta and the electron of the metal is instantaneous ( ≃ 10 - 9 s e c . ) (3). The interacting photon transfers its whole energy to a single electron. (4) The energy of photon goes in two parts: a part of energy is used in ejecting the electron from the metal surface(i.e., in overcoming work function) and the remaining part appears in the form of kinetic energy of the same electron If ν be the frequency of incident light, the energy of photon h ν If ϕ 0 be the work function, The maximum KE of the emitted electrons is K max = 1 2 mv 2 = hν - ϕ 0 If the incident photon is of the threshold frequency ( ν 0 ), just sufficient to eject the electron from the metal, then ϕ 0 = hν 0 Hence, the equation becomes 1 2 m v m a x 2 = hν - hv 0 = h ν - ν 0 or hν = W + 1 2 mv 2 max where W = work function( ϕ 0 )

Explain the Maxwell quantum theory in connection with electromagnetic radiation and derive the relation λ = h p , where λ is wavelength p is momentum of photon by using Einstein's concept of mass energy relation.

Maxwell demonstrated that electric and magnetic fields travel through space as transverse waves moving at the speed of light. He, with the help of his four equations established the that oscillating electric and magnetic fields generate each other and light is an electromagnetic wave. de Broglie concluded that a moving particle behaves as a wave and vice-versa and established a relation between them According tom the plank's quantum theory, the energy associated with the radiation of frequency ν is given by E = hν where h = Planck's constant. Also, the energy associated with the matter of mass m is given by the Einstein's mass=energy relation E = mc 2 , combining both, we have mc 2 = hν or m = hv c 2 The momentum of a photon having velocity c is p = mc = hν c 2 × c = hν c = hν c / ν = h λ or λ = h p This de Broglie relation for the wavelength associated with a moving particle

Find the frequency of a photon of energy 75 e V .( h = 6 . 63 × 10 - 34 Js

Given the energy of the photon= 75 eV The value of Planck's constant The energy of a photon is given by the Planck's relation h = 6 . 63 × 10 - 34 Js E = hν where ν = frequency of the radiation and h = Planck's constant Substituting the values in the formula, we have 75 e V = 6 . 63 × 10 - 34 × ν or ν = 75 × 1 . 6 × 10 - 19 6 . 63 × 10 - 34 = 1 . 8 × 10 16 Hz

A certain spectral line has a wavelength of 4000 A o . Calculate the frequency and energy of the photons associated with it. e = 1 . 6 × 10 - 19 C , h = 6 . 63 × 10 - 34 Js

Given the wavelength of line λ = 4000 × 10 - 10 m e = 1 . 6 × 10 - 19 C , h = 6 . 63 × 10 - 34 Js The frequency of the radiation is given by ν = c λ where c , λ are the velocity and wave length of the light. Putting these values we have the frequency of the spectral line = 3 × 10 8 4000 × 10 - 10 = 0 . 75 × 10 15 H z The energy of a photon is given by the Planck's relation E = hν where ν = frequency of the radiation and h = Planck's constant Substituting the values in the formula, we have E = 6 . 63 × 10 - 34 × 0 . 75 × 10 15 1 . 6 × 10 - 19 = 3 . 1 eV

The energy of 6000 A o photon is 3 . 30 × 10 - 19 J . Find the energy of 4000 A o . e = 1 . 6 × 10 - 19 C , h = 6 . 63 × 10 - 34 Js

Given energy of photon E = 3 . 3 × 10 - 10 J . Wavelength of light λ = 6000 × 10 - 10 m The energy of a photon is given by Planck's relation E = hν = hc λ Where ν , c a n d λ are the frequency, speed and wave length of light. It indicates that E ∝ 1 λ Therefore, we have energy of photone 4000 ° A = 6000 × energy of 6000 ° A 4000 = 3 . 3 × 10 - 19 × 1 . 5 = 4 . 95 × 10 - 19 J .

E M B I B E

ISC Physics Class 12, Part-2>Chapter 10 - Photoelectric Effect>QUESTIONS>Q 2

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12th ICSE

IMPORTANT

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Define threshold frequency, work function and cut-off potential. Draw a graph between maximum kinetic energy against frequency of light falling on a metal surface.

Important Questions on Photoelectric Effect

MEDIUM

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 10 - Photoelectric Effect>QUESTIONS>Q 3

Explain Einstein's theory in connection with photoelectric effect. Plot a graph between stopping potential against frequency of incident radiations and show that slope of the graph is ratio of Planck's constant and charge of electron.