Mathematics PHOTOELECTRIC EFFECT, WAVE THEORY OF LIGHT AND EINSTEINS PHOTOELECTRIC EQUATION

### Topic Covered

color{blue}{star} PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT
color{blue}{star} EINSTEIN’S PHOTOELECTRIC EQUATION
color{blue}{star} PARTICLE NATURE OF LIGHT: THE PHOTON

### PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT

color{blue} ✍️The wave nature of light was well established by the end of the nineteenth century. The phenomena of interference, diffraction and polarisation were explained in a natural and satisfactory way by the wave picture of light.

color{blue} ✍️According to this picture, light is an electromagnetic wave consisting of electric and magnetic fields with continuous distribution of energy over the region of space over which the wave is extended. Let us now see if this wave picture of light can explain the observations on photoelectric emission given in the previous section.

color{blue} ✍️According to the wave picture of light, the free electrons at the surface of the metal (over which the beam of radiation falls) absorb the radiant energy continuously. The greater the intensity of radiation, the greater are the amplitude of electric and magnetic fields.

color{blue} ✍️Consequently, the greater the intensity, the greater should be the energy absorbed by each electron. In this picture, the maximum kinetic energy of the photoelectrons on the surface is then expected to increase with increase in intensity.

color{blue} ✍️Also, no matter what the frequency of radiation is, a sufficiently intense beam of radiation (over sufficient time) should be able to impart enough energy to the electrons, so that they exceed the minimum energy needed to escape from the metal surface .

color{blue} ✍️A threshold frequency, therefore, should not exist. These expectations of the wave theory directly contradict observations (i), (ii) and (iii) given at the end of previous sub-section.

color{blue} ✍️Further, we should note that in the wave picture, the absorption of energy by electron takes place continuously over the entire wavefront of the radiation.

color{blue} ✍️Since a large number of electrons absorb energy, the energy absorbed per electron per unit time turns out to be small. Explicit calculations estimate that it can take hours or more for a single electron to pick up sufficient energy to overcome the work function and come out of the metal.

color{blue} ✍️This conclusion is again in striking contrast to observation (iv) that the photoelectric emission is instantaneous. In short, the wave picture is unable to explain the most basic features of photoelectric emission.

### EINSTEIN’S PHOTOELECTRIC EQUATION: ENERGY QUANTUM OF RADIATION

color{blue} ✍️Albert Einstein proposed a radically new picture of electromagnetic radiation to explain photoelectric effect. In this picture, photoelectric emission does not take place by continuous absorption of energy from radiation.

color{blue} ✍️Radiation energy is built up of discrete units – the so called quanta of energy of radiation. Each quantum of radiant energy

color {blue}{(K_(max) = hv - phi_0)}

.......(11.2)

color{blue} ✍️More tightly bound electrons will emerge with kinetic energies less than the maximum value.

color{brown} {"Note"} that the intensity of light of a given frequency is determined by the number of photons incident per second. Increasing the intensity will increase the number of emitted electrons per second. However, the maximum kinetic energy of the emitted photoelectrons is determined by the energy of each photon.

color{blue} ✍️Equation (11.2) is known as Einstein’s photoelectric equation. We now see how this equation accounts in a simple and elegant manner all the observations on photoelectric effect given at the end of previous sub-section.

color{blue} ✍️ According to Eq. (11.2), K_(max) depends linearly on n, and is independent of intensity of radiation, in agreement with observation.

color{blue} ✍️This has happened because in Einstein’s picture, photoelectric effect arises from the absorption of a single quantum of radiation by a single electron. The intensity of radiation (that is proportional to the number of energy quanta per unit area per unit time) is irrelevant to this basic process.

color{blue} ✍️ Since Kmax must be non-negative, Eq. (11.2 ) implies that photoelectric emission is possible only if color{purple}(h n > phi0)

or color{purple}(n > V_0,) where

color {blue}{(V_0 = (phi_0)/(h))}

......(11.3)

color{blue} ✍️Equation (11.3) shows that the greater the work function color{purple}(phi_0), the higher the minimum or threshold frequency color{purple}(V_0) needed to emit photoelectrons.

color{blue} ✍️Thus, there exists a threshold frequency color{purple}(V_0 = (phi_0//h)) for the metal surface, below which no photoelectric emission is possible, no matter how intense the incident radiation may be or how long it falls on the surface.

color{blue} ✍️ In this picture, intensity of radiation as noted above, is proportional to the number of energy quanta per unit area per unit time.

color{blue} ✍️The greater the number of energy quanta available, the greater is the number of electrons absorbing the energy quanta and greater, therefore, is the number of electrons coming out of the metal (for color{purple}(n > V_0)). This explains why, for color{purple}(n > V_0) , photoelectric current is proportional to intensity.

color{blue} ✍️ In Einstein’s picture, the basic elementary process involved in photoelectric effect is the absorption of a light quantum by an electron. This process is instantaneous.

color{blue} ✍️Thus, whatever may be the intensity i.e., the number of quanta of radiation per unit area per unit time, photoelectric emission is instantaneous.
Low intensity does not mean delay in emission, since the basic elementary process is the same. Intensity only determines how many electrons are able to participate in the elementary process (absorption of a light quantum by a single electron) and, therefore, the photoelectric current.

color{blue} ✍️Using Eq. (11.1), the photoelectric equation, Eq. (11.2), can be written as

color{purple}(e V_0 = h n – phi_0" for "n.= V_0)

or

color {blue}(V_0 = h/e V - (phi_0)/e)

.......(11.4)

color{blue} ✍️This is an important result. It predicts that the V_0 versus n curve is a straight line with slope = (h/e), independent of the nature of the material. During 1906-1916, Millikan performed a series of experiments on photoelectric effect, aimed at disproving Einstein’s photoelectric equation.

color{blue} ✍️He measured the slope of the straight line obtained for sodium, similar to that shown in Fig. 11.5. Using the known value of e, he determined the value of Planck’s constant h. was close to the value of Planck’s contant color{purple}((= 6.626 × 10^(–34)J s)) determined in an entirely different context. In this way, in 1916, Millikan proved the validity of Einstein’s photoelectric equation, instead of disproving it.

color{blue} ✍️The successful explanation of photoelectric effect using the hypothesis of light quanta and the experimental determination of values of h and phi_0, in agreement with values obtained from other experiments, led to the acceptance of Einstein’s picture of photoelectric effect. Millikan verified photoelectric equation with great precision, for a number of alkali metals over a wide range of radiation frequencies.

### PARTICLE NATURE OF LIGHT: THE PHOTON

color{blue} ✍️Photoelectric effect thus gave evidence to the strange fact that light in interaction with matter behaved as if it was made of quanta or packets of energy, each of energy h nu.

color{blue} ✍️Einstein arrived at the important result, that the light quantum can also be associated with momentum color{purple}((h nu//c).)

color{blue} ✍️A definite value of energy as well as momentum is a strong sign that the light quantum can be associated with a particle. This particle was later named photon. The particle-like behaviour of light was further confirmed, by the experiment of A.H.

color{blue} ✍️Compton on scattering of X-rays from electrons. Einstein was awarded the Nobel Prize in Physics for his contribution to theoretical physics and the photoelectric effect. In 1923, Millikan was awarded the Nobel Prize in physics for his work on the elementary charge of electricity and on the photoelectric effect. We can summarise the photon picture of electromagnetic radiation as follows:

color{blue} {(i)} In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.

color{blue} {(ii)} Each photon has energy color{purple}(E =(h nu)) and momentum color{purple}(p = (h nu//c),) and speed c, the speed of light.

color{blue} {(iii)} All photons of light of a particular frequency n, or wavelength l, have the same energy color{purple}(E (=h nu = hc//λ)) and momentum color{purple}(p = (h nu//c = h//λ),) whatever the intensity of radiation may be. By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.

color{blue} {(iv) } Photons are electrically neutral and are not deflected by electric and magnetic fields.

color{blue} {(v) } In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, the number of photons may not be conserved in a collision. The photon may be absorbed or a new photon may be created.
Q 3149456313

Monochromatic light of frequency 6.0 ×10^(14) Hz is produced by a laser. The power emitted is 2.0 ×10–3 W. (a) What is the energy of a photon in the light beam? (b) How many photons per second, on an average, are emitted by the source?
Class 12 Chapter 11 Example 1
Solution:

(a) Each photon has an energy

E = h n = ( 6.63 ×10^(–34) J s) (6.0 ×10^(14) Hz)

(b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = N E. Then

N = P/E = (2.0xx10^3W)/(3.98xx10^(19)J)

= 5.0 ×10^(15) photons per second.
Q 3109456318

The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.
Class 12 Chapter 11 Example 2
Solution:

(a) For the cut-off or threshold frequency, the energy h V_0 of the incident radiation must be equal to work function phi_0, so that
V_0 = (phi_0)\h= (2.14eV)/(6.63xx10^(-34)Js)

= (2.14 xx 1.6 xx10^(-19)m J)/(6.63 xx 10^(-34) Js) = 5.16 xx 10^(14) Hz

Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy e V0 by the retarding potential V_0. Einstein’s Photoelectric equation is

eV_) = hv-phi_0 = (hc)/(lamda) - phi_0

or, lamda = hc/(eV-0+phi_0)

= ((6.63xx10^(-34) Js) xx (3xx10^8 m//s))/(0.60eV+2.14eV)

= (19.89xx10^(-26)JM)/(2.74eV)

lamda= (19.89xx10^(-26)JM)/(2.74 xx1.6 xx10^(-19)J) = 454nm
Q 3139556412

The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellowgreen colour and about 760 nm for red colour.
(a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum? (Take h = 6.63×10–34 J s and 1 eV = 1.6×10 –19J.)
(b) From which of the photosensitive materials with work functions listed in Table 11.1 and using the results of (i), (ii) and (iii) of (a), can you build a photoelectric device that operates with visible light?

Solution:

(a) Energy of the incident photon, E = hn = hc//lamda
E = (6.63×10^(–34)J s) (3×10^8 m//s)/lamda

(i) For violet light, lamda_1 = 390 nm (lower wavelength end)

Incident photon energy, E_1 = (1.989xx10^(-25)Jm)/(390xx10^(-9)m)
= 5.10 × 10^(–19)J

(= 5.10 × 10^(–19)J)/(1.6xx10^(-19)/J/eV)
= 3.19 eV

(ii) For yellow-green light, lamda_2= 550 nm (average wavelength)

Incident photon energy, E_2 (1.989xx10^(-25)Jm)/(550xx10^(-9)m)

= 3.62×10^(–19) J = 2.26 eV

(iii) For red light, lamda_3 = 760 nm (higher wavelength end)
Incident photon energy, E_2 (1.989xx10^(-25)Jm)/(760xx10^(-9)M)

= 2.62×10^(–19) J = 1.64 eV

(b) For a photoelectric device to operate, we require incident light energy E to be equal to or greater than the work function phi_0 of the material. Thus, the photoelectric device will operate with violet light (with E = 3.19 eV) photosensitive material Na (with phi_0 = 2.75 eV), K (with phi_0 = 2.30 eV) and Cs (with phi_0 = 2.14 eV). It will also operate with yellow-green light (with E = 2.26 eV) for Cs (with phi_0 = 2.14 eV) only. However, it will not operate with red light (with E = 1.64 eV) for any of these photosensitive materials.