Dual Nature of Radiation

The dual nature of radiation is the concept in quantum mechanics that all matter and energy have both wave-like and particle-like properties. For radiation, this means that light, which we often described as a wave, can also behave as a particle under certain conditions. The particle of light is called a photon.

Dual Nature of Radiation Definition

The dual nature of radiation refers to the property of light and electromagnetic waves to exhibit both wave-like properties (interference and diffraction) and particle-like properties (photons with quantized energy).

1.0Particle Nature of Light: The Photon

Photon Theory of Light

We have learnt that light has wave character as well as particle character. Depending on the situation, one of the two characters dominates. When light is passed through a double slit, it shows interference. This observation can only be understood in terms of wave theory which was discussed in detail in an earlier chapter. There are some phenomena which can only be understood in terms of the particle theory of light. When light of sufficiently low wavelength falls on a metal surface, electrons are ejected. This phenomenon is called the photoelectric effect and can be understood only in terms of the particle nature of light. The particles of light have several properties in common with the material particles and several other properties which are different from the material particles. The particles of light are called photons. We list some of the important properties of photons.

1. A photon always travels at a speed c = 299,792,458 ms⁻¹ ≈ 3 × 10⁸ ms⁻¹ in a vacuum. This is true for any frame of reference used to observe the photon.
2. The mass of a photon is not defined in the sense of Newtonian mechanics. We shall ignore this concept. We simply state that the rest mass of a photon is zero.
3. Each photon has a definite energy and a definite linear momentum.
4. Let E and p be the energy and linear momentum of a photon of light, and and be the frequency and wavelength of the same light when it behaves as a wave. Then,
   And  E = hν = hc/λ    
   p = h/λ = E/c
   
   where h is a universal constant known as the Planck constant and has a value = 4.136 × 10⁻³⁴ eV s  (h = 6.6 × 10⁻³⁴ J s)
   Thus, all photons of light of a particular wavelength λ have the same energy $E=\frac{hc}{\lambda }$and the same momentum $p=\frac{h}{\lambda }$

5. A photon may collide with a material particle. The total energy and the total momentum remain conserved in such a collision. The photon may get absorbed and/or a new photon may be emitted. Thus, the number of photons may not be conserved.

6.If the intensity of light of a given wavelength is increased, there is an increase in the number of photons crossing a given area in a given time. The energy of each photon remains the same.

2.0Wave Nature of Light

For centuries, light was understood as a wave. This wave nature is beautifully demonstrated by phenomena that a particle model cannot explain. Key proofs of light's wave nature include:

• Diffraction: The bending of light around obstacles or through narrow apertures.
• Interference: The superposition of two or more waves to form a new wave pattern, resulting in alternating bright and dark fringes.
• Polarization: The restriction of light wave vibrations to a single plane.

The existence of both sets of phenomena led physicists to accept the dual nature of light.

3.0The Photoelectric Effect

When electromagnetic radiation of suitable wavelength is incident on a metallic surface then electrons are emitted, this phenomenon is called photo electric effect.

Photoelectron: The electron emitted in photoelectric effect is called photoelectron.

Photoelectric current: If current passes through the circuit in photoelectric effect then the current is called photoelectric current.

Work function: The minimum energy required to make an electron free from the metal is called work function. It is constant for a metal and denoted by $\varphi$ or W. It is the minimum for Cesium. It is relatively less for alkali metals.

Work functions of some photosensitive metals

To produce photo electric effect only metal and light is necessary but for observing it, the circuit is completed. Figure shows an arrangement used to study the photoelectric effect.                 

Here the plate (1) is called emitter or cathode and other plate (2) is called collector or anode.

Saturation current: When all the photo electrons emitted by cathode reach the anode then current flowing in the circuit at that instant is known as saturation current, this is the maximum value of photoelectric current.

Stopping potential: Minimum magnitude of negative potential of anode with respect to cathode for which current is zero is called stopping potential. This is also known as cut off voltage. This voltage is independent of intensity. 

Retarding potential: Negative potential of anode with respect toHere the plate (1) is called emitter or cathode and other plate (2) is called collector or anode.

Observations made by Einstein: 

A graph between intensity of light and photoelectric current is found to be a straight line as shown in figure. Photoelectric current is directly proportional to the intensity of incident radiation. In this experiment the frequency and retarding potential are kept constant. A graph between photoelectric current and potential difference between cathode and anode is found as shown in figure.                             

In case of saturation current, 

Rate of emission of photoelectrons = rate of flow of photoelectrons, 

Here, vₛ → stopping potential and it is a positive quantity

Electrons emitted from the surface of metal have different energies.

Maximum kinetic energy of photoelectron on the cathode = eVₛ

${\text{K.E}}_{\max }=e{V}_s$

Whenever photoelectric effect takes place, electrons are ejected out with kinetic energies ranging from $\text{0 to K.E}{(}_{\max })$ i.e. 0 ≤ K.E(_e) ≤ eVₛ 

The energy distribution of photoelectrons is shown in figure.

If intensity is increased (keeping the frequency constant) then saturation current is increased by the same factor by which intensity increases. Stopping potential is the same, so maximum value of kinetic energy is not affected. 

If light of different frequencies is used then obtained plots are shown in figure. 

It is clear from the graph, as v increases, stopping potential increases, it means the maximum value of kinetic energy increases. 

Graphs between maximum kinetic energy of electrons ejected from different metals and frequency of light used are found to be straight lines of the same slope as shown in figure below.

  Graph between Kₘₐₓ and ν.

m₁, m₂, m₃ : Three different metals

 It is clear from the graph that there is a minimum frequency of electromagnetic radiation which can produce a photoelectric effect, which is called threshold frequency.

νₜₕ = Threshold Frequency

For Photoelectric Effect

ν ≥ νₜₕ

For Photoelectric Effect

ν < νₜₕ

Minimum frequency For Photoelectric Effect = νₜₕ

νₘᵢₙ = νₜₕ

Threshold wavelength (λₜₕ) → The maximum wavelength of radiation which can produce photoelectric effect.

λ ≤ λₜₕ  for photo electric effect

Maximum wavelength for photoelectric effect λₘₐₓ = λₜₕ

Now writing an equation of a straight line from a graph.

We have Kₘₐₓ = Aν + B

When ν = νₜₕ , Kₘₐₓ = 0 and B = −Aνₜₕ

Hence [ Kₘₐₓ = A(ν − νₜₕ) ] and 
A = tan θ = 6.6 × 10⁻³⁴ J·s (from experimental data)

Later on ‘A’ was found to be ‘h’. It is also observed that the photoelectric effect is an instantaneous process. When light falls on the surface electrons start ejecting without taking any time.

Three major features of the photoelectric effect cannot be explained in terms of the classical wave theory of light. 

Intensity: The energy crossing per unit area per unit time perpendicular to the direction of propagation is called the intensity of a wave.

Consider a cylindrical volume with area of cross-section A and length c t along the X-axis. The energy contained in this cylinder crosses the area A in time t as the wave propagates at speed c. The energy contained.

$$\begin{gathered} U=u_{av}(c\Delta t)A \\ \text{The intensity is} \\ I=\frac{U}{A\Delta t}=u_{av}c \\ \text{In the terms of maximum electric field,} \\ I=\frac{1}{2}{\epsilon }_0{E}_0^{2}c \\ \text{If we consider light as a wave then the intensity depends upon the electric field.} \\ \text{If we take work function} \\ W=I.A.t\text{then}t=\frac{W}{IA} \\ \text{So, for photoelectric effect there should be time lag because the metal has a work function. But it is observed that the photoelectric effect is an instantaneous process. Hence, light is not of wave nature.} \end{gathered}$$

The Intensity Problem : Wave theory requires that the oscillating electric field vector Eof the light wave increases in amplitude as the intensity of the light beam is increased. Since the force applied to the electron is eE, this suggests that the kinetic energy of the photoelectrons should also increase as the light beam is made more intense. However, observation shows that maximum kinetic energy is independent of the light intensity

The Frequency Problem : According to the wave theory, the photoelectric effect should occur for any frequency of the light, provided only that the light is intense enough to supply the energy needed to eject the photoelectrons. However, observations shows that there exists for each surface a characteristic cut off frequency νₜₕ, for frequencies less than νₜₕ the photoelectric effect does not occur, no matter how intense the light beam.

The Time Delay Problem : If the energy acquired by a photoelectron is absorbed directly from the wave incident on the metal plate, the "effective target area" for an electron in the metal is limited and probably not much more than that of a circle of diameter roughly equal to that of an atom. In the classical theory, the light energy is uniformly distributed over the wave front. Thus, if the light is feeble enough, there should be a measurable time lag, between the impinging of the light on the surface and the ejection of the photoelectron. During this interval the electron should be absorbing energy from the beam until it had accumulated enough to escape. However, no detectable time lag has ever been measured. Now, quantum theory solves these problems in providing the correct interpretation of the photoelectric effect.

Planck's Quantum Theory

The light energy from any source is always an integral multiple of a smaller energy value called quantum of light. Hence energy,Q = NE,

where E = hν and N (number of photons) = 1, 2, 3, …

Here energy is quantized. h is the quantum of energy, it is a packet of energy called a photon.

$E=h\nu =\frac{hc}{\lambda }andhc=12400\text{eV}\text{A˚}$

Einstein's Photon Theory

In 1905 Einstein made a remarkable assumption about the nature of light that, under some circumstances,it behaves as if its energy is concentrated into localized bundles, later called photons. The energy E of a single photon is given by

E = hν

If we apply Einstein's photon concept to the photoelectric effect, we can write

$h\nu =W+K_{\max }\text{(energy conservation)}$

Equation says that a single photon carries an energy hν into the surface where it is absorbed by a single electron. Part of this energy W (called the work function of the emitting surface) is used in causing the electron to escape from the metal surface. The excess energy (hν − W) becomes the electron's kinetic energy; if the electron does not lose energy by internal collisions as it escapes from the metal, it will still have this much kinetic energy after it emerges. Thus, Kₘₐₓ represents the maximum kinetic energy that the photoelectron can have outside the surface. There is complete agreement of the photon theory with experiment.

Now IA = Nhν

⇒ N = IA / hν= number of photons incident per unit time on an area 'A' when light of intensity 'I' is incident normally.

If we double the light intensity, we double the number of photons and thus double the photoelectric current; we do not change the energy of the individual photons or the nature of the individual photoelectric processes.

The second objection (the frequency problem) is met if Kₘₐₓ equals zero, we have

hνₜₕ = W

Which asserts that the photon has just enough energy to eject the photoelectrons and none extra to appear as kinetic energy. If v is reduced below νₜₕ hv will be smaller than W and the individual photons, no matter how many of them there are (that is, no matter how intense the illumination), will not have enough energy to eject photoelectrons.

 The third objection (the time delay problem) follows from the photon theory because the required energy is supplied in a concentrated bundle. It is not spread uniformly over the beam cross section as in the wave theory. Hence Einstein's equation for photoelectric effect is given by  

$h\nu =h{\nu }_{th}+K_{\max };K_{\max }=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{th}}$

De-Broglie Wavelength of Matter Wave

$$\begin{gathered} \text{A photon of frequency and wavelength has energy.} \\ E=h\nu =\frac{hc}{\lambda } \\ \text{By Einstein’s energy mass relation,}E=m{c}^2\text{ the equivalent mass m of the photon is given by,} \\ m=\frac{E}{c^2}=\frac{h\nu }{c^2}=\frac{h}{\lambda c}.........(1) \\ \lambda =\frac{h}{mc}\text{or}\lambda =\frac{h}{p}........(2) \\ \text{Here p is the momentum of a photon. By analogy de-Broglie suggested that a particle of mass m moving with speed v behaves in some ways like waves of wavelength (called de-Broglie wavelength and the wave is called matter wave) given by,} \\ \lambda =\frac{h}{mv}=\frac{h}{p}........(3) \\ \text{Where p is the momentum of the particle. Momentum is related to the kinetic energy by the equation} \\ p=\sqrt{2Km} \\ \text{and a charge q when accelerated by a potential difference V gains a kinetic energyK = qV. Combining all these relations equation (iii), can be written as,} \\ \lambda =\frac{h}{mv}=\frac{h}{p}=\frac{h}{\sqrt{2Km}}=\frac{h}{\sqrt{2qVm}}(\text{de-Broglie wavelength}).........(4) \end{gathered}$$

De-Broglie Wavelength for an Electron:

If an electron (charge = e)is accelerated by a potential of V  volts, it acquires a kinetic energy,

K = eV

Substituting the values of h, m and q In equation (4), we get a simple formula for calculating de-Broglie wavelengths of an electron.

$\lambda (\text{in A˚})=\sqrt{\frac{150}{V(\text{in volts})}}$

De-Broglie Wavelength of a Gas Molecule

Let us consider a gas molecule at absolute temperature T. Kinetic energy of gas molecule is given by

$$\begin{gathered} \text{K.E.}=\frac{3}{2}kT;k=\text{Boltzmann constant} \\ \therefore {\lambda }_{\text{gas molecule}}=\frac{h}{\sqrt{3mkT}} \end{gathered}$$

4.0Solved Problems

Illustration-1: Find the ratio of De-Broglie wavelength of molecules of hydrogen and helium which are at temperatures 27° C and 127° C respectively.

Solution: De-Broglie wavelength is given by

$\therefore \frac{{\lambda }_{H_2}}{{\lambda }_{He}}=\sqrt{\frac{m_{He}{T}_{He}}{m_{H_2}{T}_{H_2}}}=\sqrt{\frac{4}{2}\cdot \frac{(127+273)}{(27+273)}}=\sqrt{\frac{8}{3}}$

Illustration-2:An electron is accelerated by a potential difference of 50 volts. Find the de-Broglie wavelength associated with it.

Solution:For an electron De-Broglie wavelength is given by

$\lambda =\sqrt{\frac{150}{V}}=\sqrt{\frac{150}{50}}=\sqrt{3}=1.73\text{A˚}$

Illustration-3:Light described at a place by the equation$E=(100\text{V/m})[\sin (5\times 10^{15}{\text{s}}^{-1})+\sin (8\times 10^{15}{\text{s}}^{-1})]$ falls on a¹² metal surface having work function $2.0\text{eV}$. Calculate the maximum kinetic energy of the photoelectrons.

Solution:

$$\begin{gathered} \text{The light contains two different frequencies. The one with larger frequency will cause photoelectrons with largest kinetic energy. This larger frequency is} \\ \nu =\frac{\omega }{2\pi }=\frac{8\times 10^{15}{\text{s}}^{-1}}{2\pi } \\ \text{The maximum kinetic energy of the photoelectrons is} \\ {K}_{\max }=h\nu -W=(4.14\times 10^{-15}\text{eV - s})\left(\frac{8\times 10^{15}{\text{s}}^{-1}}{2\pi }\right)-2.0\text{eV} \\ =5.27\text{eV}-2.0\text{eV}=3.27\text{eV} \end{gathered}$$

Illustration-4: A photocell is operating in saturation mode with a photocurrent 4.8 μA when a monochromatic radiation of wavelength 3000 Å and power 1 mW is incident. When another monochromatic radiation of wavelength 1650 Å and power 5 mW is incident, it is observed that maximum velocity of photoelectron increases to two times. Assuming efficiency of photoelectron generation per incident to be the same for both the cases, calculate,

(a) threshold wavelength for the cell 

(b) efficiency of photoelectron generation. $[(\text{Number of photoelectrons emitted per incident photon})\times 100]$

 (c)saturation current in second case

Solution:

$$\begin{gathered} (a)K_1=\frac{12400}{3000}-W=4.13-W(1) \\ {K}_2=\frac{12400}{1650}-W=7.51-W(2) \\ \text{Since}{v}_2=2v_1 \\ {K}_2=4K_1(3) \\ \text{Solving above equations we get} \\ W=3\text{eV} \\ \text{Threshold Wavelength}{\lambda }_0=\frac{12400}{3}\approx 4133\text{A˚} \\ (b)\text{Energy of a photon in first case}=\frac{12400}{3000}=4.13\text{eV} \\ \text{Rate of incident photons (number of photons per second)} \\ \frac{P_1}{E_1}=\frac{10^{-3}}{6.6\times 10^{-19}}=1.5\times 10^{15}\text{per second} \\ \text{Number of electrons ejected}=\frac{4.8\times 10^{-6}}{1.6\times 10^{-19}}=3.0\times 10^{13}\text{per second} \\ \therefore \text{Efficiency of photoelectron generation} \\ \eta =\frac{3\times 10^{13}}{1.5\times 10^{15}} \\ (c)\text{Energy of photon in second case} \\ {E}_2=\frac{12400}{1650}=7.51\text{eV}=12\times 10^{-19}\text{J} \\ \text{Therefore number of photons incident per second} \\ {n}_2=\frac{P_2}{E_2}=\frac{5\times 10^{-3}}{12\times 10^{-19}}=4.17\times 10^{15}\text{per second} \\ \text{Number of electrons emitted per second} \\ =\frac{2}{100}\times 4.7\times 10^{15}=9.4\times 10^{13}\text{per second} \\ \therefore \text{saturation current in second case} \\ i=(9.4\times 10^{13})(1.6\times 10^{-19})\text{amp}=15\mu \text{A} \end{gathered}$$