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CalibratePhysics

Chapter 14

Dual Nature of Radiation and Matter

Maharashtra HSC · 6 subtopics · textbook pages 306–323

Formula SheetEvery formula this chapter can be examined on, in one table.

Constants, units and conversions the board expects

FormulaSymbols, units and conditions
h=6.63×10−34 J sc=3×108 m s−1e=1.6×10−19 Ch = 6.63 \times 10^{-34}\ \text{J s} \qquad c = 3 \times 10^{8}\ \text{m s}^{-1} \qquad e = 1.6 \times 10^{-19}\ \text{C}
  • hh is Planck's constant, cc the speed of electromagnetic radiation in vacuum, ee the magnitude of the electronic charge. Use these values unless the question prints its own.
me=9.11×10−31 kgmp=1.673×10−27 kg=1836 mem_e = 9.11 \times 10^{-31}\ \text{kg} \qquad m_p = 1.673 \times 10^{-27}\ \text{kg} = 1836\, m_e
  • Rest mass of the electron and of the proton. The ratio mp/me=1836m_p/m_e = 1836 is the one quoted in every ratio numerical.
1 eV=1.6×10−19 J1 angstrom=10−10 m1 nm=10−9 m1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J} \qquad 1\ \text{angstrom} = 10^{-10}\ \text{m} \qquad 1\ \text{nm} = 10^{-9}\ \text{m}
  • Divide an energy in joule by 1.6×10−191.6\times10^{-19} to get electron volt; multiply eV by 1.6×10−191.6\times10^{-19} to get joule. Work functions are always quoted in eV.
λ(in angstrom)×E(in eV)≈12500λ(in nm)×E(in eV)≈1250\lambda(\text{in angstrom}) \times E(\text{in eV}) \approx 12500 \qquad \lambda(\text{in nm}) \times E(\text{in eV}) \approx 1250
  • Textbook shortcut for photon energy. Useful as a quick check, not as a substitute for E=hc/λE = hc/\lambda in a full-marks answer.

Planck's Quantization of Energy

FormulaSymbols, units and conditions
E=nhνE = nh\nu
  • Energy emitted by an atomic oscillator of frequency ν\nu comes only in whole packets, n=1,2,3,…n = 1, 2, 3, \dots. This is Planck's hypothesis, the starting point of the chapter.

Photon Energy and Photon Number

FormulaSymbols, units and conditions
E=hν=hcλE = h\nu = \dfrac{hc}{\lambda}
  • Einstein's relation: energy of one photon (or quantum) of radiation of frequency ν\nu and wavelength λ\lambda. Energy in joule when hh, cc, λ\lambda are in SI. Note E∝νE \propto \nu and E∝1/λE \propto 1/\lambda.
ν=cλνˉ=1λ=Ehc\nu = \dfrac{c}{\lambda} \qquad \bar\nu = \dfrac{1}{\lambda} = \dfrac{E}{hc}
  • Frequency from wavelength; νˉ\bar\nu is the wave number in m−1\text{m}^{-1}, used when the question asks for "wave number of a photon of energy E".
n=EtotalhνN=Phν=Pλhcn = \dfrac{E_{\text{total}}}{h\nu} \qquad N = \dfrac{P}{h\nu} = \dfrac{P\lambda}{hc}
  • nn is the number of photons in a beam carrying total energy EtotalE_{\text{total}}; NN is the number of photons emitted per second by a source of power PP watt. For equal power, longer wavelength (red) means more photons per second.
Incident power=Intensity×Area\text{Incident power} = \text{Intensity} \times \text{Area}
  • Intensity in W m−2\text{W m}^{-2}, area in m2\text{m}^{2}. Used in the wave-theory time-lag estimate.

Work Function, Threshold and Einstein's Photoelectric Equation

FormulaSymbols, units and conditions
ϕ0=hν0=hcλ0\phi_0 = h\nu_0 = \dfrac{hc}{\lambda_0}
  • ϕ0\phi_0 is the work function, ν0\nu_0 the threshold frequency, λ0\lambda_0 the threshold wavelength of the emitter metal. Photoemission occurs only if ν>ν0\nu > \nu_0, i.e. λ<λ0\lambda < \lambda_0.
K.E.max=hν−ϕ0K.E._{max} = h\nu - \phi_0
  • Einstein's photoelectric equation. Write it in exactly this form when the board asks you to "state" it.
K.E.max=h(ν−ν0)=hc(1λ−1λ0)=hc(λ0−λλλ0)K.E._{max} = h(\nu - \nu_0) = hc\left(\dfrac{1}{\lambda} - \dfrac{1}{\lambda_0}\right) = hc\left(\dfrac{\lambda_0 - \lambda}{\lambda\lambda_0}\right)
  • Equivalent working forms. The last form is the quickest when both wavelengths are given.
K.E.max=eV0soV0=K.E.maxeK.E._{max} = eV_0 \qquad \text{so} \qquad V_0 = \dfrac{K.E._{max}}{e}
  • V0V_0 is the stopping (cut-off) potential in volt. If K.E.maxK.E._{max} is expressed in eV, then V0V_0 has the same number in volt.
eV0=hν−ϕ0⇒V0=(he)ν−ϕ0eeV_0 = h\nu - \phi_0 \qquad \Rightarrow \qquad V_0 = \left(\dfrac{h}{e}\right)\nu - \dfrac{\phi_0}{e}
  • Straight line of V0V_0 against ν\nu: slope =h/e= h/e (same for all metals), intercept on the V0V_0 axis =−ϕ0/e= -\phi_0/e, intercept on the ν\nu axis =ν0= \nu_0. From an experimental slope, h=e×slopeh = e \times \text{slope}.
V0=hceλ−ϕ0eV_0 = \dfrac{hc}{e\lambda} - \dfrac{\phi_0}{e}
  • Same line written against wavelength: V0V_0 varies linearly with 1/λ1/\lambda, i.e. inversely with λ\lambda.
K.E.max=12mvmax2⇒vmax=2 K.E.maxmeK.E._{max} = \dfrac{1}{2}mv_{max}^{2} \qquad \Rightarrow \qquad v_{max} = \sqrt{\dfrac{2\,K.E._{max}}{m_e}}
  • Maximum speed of the emitted photoelectron. Put K.E.maxK.E._{max} in joule, never in eV, before taking the square root.
r=mvmaxeB⇒B=mvmaxer=2m K.E.maxerr = \dfrac{m v_{max}}{eB} \qquad \Rightarrow \qquad B = \dfrac{m v_{max}}{er} = \dfrac{\sqrt{2m\,K.E._{max}}}{er}
  • Radius of the circular path of the most energetic photoelectron shot perpendicular to a magnetic field BB (tesla). Used in the exercise that asks for BB from a given radius.
I∝Intensity(ν, V fixed)I \propto \text{Intensity} \qquad (\nu,\ V \text{ fixed})
  • Photocurrent, and hence the saturation current I0I_0, is directly proportional to the intensity of the incident radiation. Doubling the intensity doubles the saturation current.

Momentum of a Photon and Compton Shift

FormulaSymbols, units and conditions
p=Ec=hνc=hλp = \dfrac{E}{c} = \dfrac{h\nu}{c} = \dfrac{h}{\lambda}
  • Momentum of a photon, valid for a massless particle moving with speed cc. Rest mass of a photon is zero; E=pcE = pc holds for a photon, not for an electron.
Δλ=λ′−λ=hmec (1−cos⁡θ)\Delta\lambda = \lambda' - \lambda = \dfrac{h}{m_e c}\,(1 - \cos\theta)
  • Compton shift for X-rays scattered through angle θ\theta by an electron. The shift depends only on θ\theta, not on the incident wavelength. Quoted result only; no numerical is set on it.

De Broglie Hypothesis

FormulaSymbols, units and conditions
λ=hp=hmvν=Eh\lambda = \dfrac{h}{p} = \dfrac{h}{mv} \qquad \nu = \dfrac{E}{h}
  • De Broglie relation for a material particle of mass mm moving with speed vv. λ\lambda is the de Broglie wavelength in metre. Greater the momentum, shorter the wavelength.
EK=p22m⇒p=2mEK⇒λ=h2mEKE_K = \dfrac{p^{2}}{2m} \qquad \Rightarrow \qquad p = \sqrt{2mE_K} \qquad \Rightarrow \qquad \lambda = \dfrac{h}{\sqrt{2mE_K}}
  • Wavelength in terms of kinetic energy. EKE_K must be in joule.
λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}
  • For a particle of charge qq accelerated from rest through a potential difference VV volt, since EK=qVE_K = qV. Valid for electron, proton or any charged ion; fails when VV is so large (kV upwards) that the speed approaches cc.
λ=1.228V(in volt) nm=12.28V(in volt) angstrom\lambda = \dfrac{1.228}{\sqrt{V(\text{in volt})}}\ \text{nm} = \dfrac{12.28}{\sqrt{V(\text{in volt})}}\ \text{angstrom}
  • The standard numerical form for an electron accelerated through VV volt. This is Eq. (14.7) of the textbook and is the fastest route in almost every de Broglie numerical.
v=2EKmv = \sqrt{\dfrac{2E_K}{m}}
  • Speed of a particle from its kinetic energy, needed when a numerical asks for wavelength, momentum and speed together.

Standard De Broglie Comparisons

FormulaSymbols, units and conditions
Same speed:λ1λ2=m2m1\text{Same speed:}\qquad \dfrac{\lambda_1}{\lambda_2} = \dfrac{m_2}{m_1}
  • Electron and proton at the same speed: λe/λp=mp/me=1836\lambda_e/\lambda_p = m_p/m_e = 1836, so the electron has the longer wavelength.
Same kinetic energy:λ1λ2=m2m1\text{Same kinetic energy:}\qquad \dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{m_2}{m_1}}
  • Electron and proton of equal kinetic energy: λe/λp=1836=42.85\lambda_e/\lambda_p = \sqrt{1836} = 42.85, electron longer. Of electron, proton, alpha particle and hydrogen atom at equal EKE_K, the electron has the longest wavelength because it is the lightest.
Same momentum:λ1=λ2\text{Same momentum:}\qquad \lambda_1 = \lambda_2
  • Equal momenta means equal wavelengths, whatever the masses.
Same wavelength:p1=p2andE1E2=m2m1\text{Same wavelength:}\qquad p_1 = p_2 \quad \text{and} \quad \dfrac{E_1}{E_2} = \dfrac{m_2}{m_1}
  • Electron and proton of equal de Broglie wavelength: momenta ratio 11, kinetic energy ratio Ee/Ep=mp/me=1836E_e/E_p = m_p/m_e = 1836.

Davisson and Germer Experiment

FormulaSymbols, units and conditions
V=54 Vθ=50∘λtheory=1.22854=0.167 nmV = 54\ \text{V} \qquad \theta = 50^\circ \qquad \lambda_{\text{theory}} = \dfrac{1.228}{\sqrt{54}} = 0.167\ \text{nm}
  • Accelerating potential and scattering angle at which the intensity peak was found, and the de Broglie wavelength that follows from Eq. (14.7).
λexperiment=0.165 nm\lambda_{\text{experiment}} = 0.165\ \text{nm}
  • Wavelength obtained from the electron-diffraction measurement. Close agreement of 0.1670.167 nm with 0.1650.165 nm is what verifies the de Broglie relation.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.6 subtopics
  1. 1Introduction
  2. 2The Photoelectric Effect24 board questions
  3. 3Wave-Particle Duality of Electromagnetic Radiation
  4. 4Photo Cell2 board questions
  5. 5De Broglie Hypothesis8 board questions
  6. 6Davisson and Germer Experiment1 board question
The whole chapter in brief

14.1 Introduction

  • Reflection, refraction, interference, diffraction and polarization are explained by treating light as an electromagnetic wave: mutually perpendicular oscillating electric and magnetic fields, both perpendicular to the direction of propagation.
  • Black body radiation could not be explained by the wave picture. Such phenomena appear when radiation interacts with matter and need quantum physics.
  • Planck's quantization of energy: energy is emitted in packets, and at higher frequency the packet is larger. Atoms behave like tiny oscillators emitting radiation only in discrete packets E=nhνE = nh\nu, emission occurring when the oscillator jumps from one quantized level to a lower one.
  • Planck's model became the basis of Einstein's explanation of the photoelectric effect.

14.2 The Photoelectric Effect

  • Photoelectric effect: the phenomenon of emission of electrons from a metal surface when radiation of appropriate frequency is incident on it.
  • A surface that emits electrons when illuminated with appropriate radiation is a photosensitive surface. The emitted electrons are called photoelectrons (they are ordinary electrons; the name only records how they were released).
  • Discovered by Heinrich Hertz in 1887 while producing electromagnetic waves by spark discharge.
  • Zinc, cadmium and magnesium need ultraviolet radiation; alkali metals respond even to visible light.
  • Two ways of getting electrical energy from light: photo-emissive effect (electrons ejected from a surface, as here) and photo-voltaic effect (a potential difference generated in a semiconductor, as in a solar cell).

Experimental set-up

  • Evacuated glass tube with a quartz window, containing a photosensitive emitter plate E (cathode) and a collector plate C (anode).
  • E and C are connected to a variable voltage source and an ammeter; a voltmeter reads the potential difference VV between them.
  • Positive anode potential accelerates the photoelectrons (accelerating potential); negative anode potential retards them (retarding potential).
  • Source S gives monochromatic light of sufficiently high frequency, wavelength ≤10−7\leq 10^{-7} m.
  • The quartz window matters: ordinary glass absorbs ultraviolet radiation.

The nine observations

  • Photocurrent flows only if ν>ν0\nu > \nu_0, the threshold frequency, which is fixed for a given metal and different for different metals. Above ν0\nu_0 electrons are emitted however feeble the light.
  • Emission is instantaneous: the current starts within about 10−910^{-9} s and stops the moment the radiation is cut off. There is no time lag.
  • For fixed ν\nu and fixed accelerating potential VV, the photocurrent increases linearly with intensity.
  • For fixed incident radiation, the current first rises with the accelerating potential and then becomes constant at the saturation current I0I_0.
  • Increasing the intensity increases I0I_0 proportionately: double the intensity, double the saturation current.
  • K.E.maxK.E._{max} of the photoelectrons depends on the emitter material and on the incident frequency, and is completely independent of the intensity.
  • Increasingly negative collector potential reduces the current to zero at −V0-V_0, the cut-off or stopping potential. V0V_0 is the same for any intensity at a given frequency, but different for different emitter materials.
  • Changing ν\nu at constant intensity leaves the saturation current unchanged but changes V0V_0; V0V_0 varies linearly with ν\nu.
  • The graph of V0V_0 against ν\nu is a straight line whose slope is the same for all metals but whose intercepts differ.
  • Photocurrent, and hence the number of electrons, depends on intensity and not on frequency, so long as ν>ν0\nu > \nu_0 and the anode is positive.

Work function

  • Work function ϕ0\phi_0: the minimum amount of energy required to be provided to an electron to pull it out of the metal from the surface.
  • It depends on the metal and on the nature of its surface, and is quoted in eV. What ionization energy is to an atom, the work function is to a solid.
MetalWork function (eV)MetalWork function (eV)
Potassium2.3Silver4.3
Sodium2.4Aluminium4.3
Calcium2.9Tungsten4.5
Zinc3.6Copper4.7
Nickel5.0Gold5.1
  • Highest work function in the table is gold, so gold needs the highest incident frequency to generate photocurrent. Potassium has the lowest, so it works with visible light and is the cathode material used in a street-light photocell.

Failure of the wave theory

ObservationWave theory predictsPhoton picture explains
Electrons emitted as soon as light is incidentVery intense light needed for instantaneous emissionOne photon ejects one electron; the exchange on collision is instantaneous
Very low intensity is still sufficientLow intensity should give no photocurrentLow intensity means fewer photons, not weaker photons, so a small current flows
Higher intensity gives larger photocurrentHigher intensity means higher energy radiationMore photons per unit time means more electrons per unit time
Intensity has no effect on electron energyHigher intensity should give higher electron energyPhoton energy is fixed by frequency and does not depend on intensity
A threshold frequency is neededLow frequency should still work, only more slowlyA low-frequency photon has too little energy to free an electron
Higher frequency raises K.E.maxK.E._{max}Only intensity should matter, not frequencyHigher frequency means a more energetic photon, so more energy is left over as kinetic energy
  • Numerical estimate in the textbook: for potassium of area 5 cm25\ \text{cm}^2 under intensity 0.5×10−4 W m−20.5\times10^{-4}\ \text{W m}^{-2}, the wave picture gives a time lag of about 4.428×1044.428\times10^{4} s, roughly half a day, against an observed lag of under 10−910^{-9} s.

Einstein's explanation

  • Einstein (1905) extended Planck's quantization to all electromagnetic radiation. A quantum of light is a photon of energy E=hνE = h\nu.
  • The photon gives all its energy to one electron and ceases to exist. Emission occurs only if hν≥ϕ0h\nu \geq \phi_0, giving ν0=ϕ0/h\nu_0 = \phi_0/h.
  • Energy left over after escaping is the kinetic energy: K.E.max=hν−ϕ0K.E._{max} = h\nu - \phi_0, Einstein's photoelectric equation.
  • Emitted electrons have a spread of energies because electrons in a solid do not all have the same energy, are ejected from different depths, and may suffer collisions on the way out. K.E.maxK.E._{max} belongs to the most energetic one.
  • With K.E.max=eV0K.E._{max} = eV_0, the equation becomes V0=(h/e)ν−ϕ0/eV_0 = (h/e)\nu - \phi_0/e, which is why all the V0V_0 against ν\nu lines have the same slope h/eh/e and different intercepts.
  • Millikan's 1909 measurement of ee, and of hh from this slope, matched Planck's value and settled the matter. Einstein received the Nobel Prize in 1921 and Millikan in 1923.

14.3 Wave-Particle Duality of Electromagnetic Radiation

  • Compton (1923-24) sent monochromatic X-rays of wavelength λ\lambda onto graphite and found the scattered intensity peaked at two wavelengths: the original λ\lambda and a longer λ′\lambda'.
  • He treated it as a billiard-ball collision of an X-ray photon with an electron. Energy and momentum are transferred, so the scattered photon has lower energy, lower frequency and longer wavelength.
  • Compton shift Δλ=hmec(1−cos⁡θ)\Delta\lambda = \dfrac{h}{m_e c}(1 - \cos\theta), depending only on the scattering angle θ\theta and not on the incident wavelength. Letting h→0h \to 0 recovers the wave-theory result.
  • Photon facts: a photon carries energy and momentum; all photons of a given frequency have the same energy and momentum; photons are electrically neutral and are not deflected by electric or magnetic fields; in a photon-particle collision energy and momentum are conserved but the number of photons is not; a photon always moves with speed cc and is never at rest; its rest mass is zero in all frames.
  • Wave nature shows in interference and diffraction, when slit width or slit separation is smaller than or comparable to the wavelength. Particle nature shows in black body radiation, the photoelectric effect and Compton scattering.
  • Since neither picture alone covers all the observations, both must be kept: one dominates in some situations and the other in the rest. This is the wave-particle duality of electromagnetic radiation.

14.4 Photo Cell

  • A photo cell is a device that uses the photoelectric effect to convert light energy into electrical energy.
  • Construction: a semi-cylindrical photosensitive metal plate E as cathode and a wire loop collector C as anode, supported in an evacuated glass or quartz bulb, connected to an external circuit with a high tension battery B and a microammeter. The photosensitive material may instead be pasted as a thin film on the inner wall of the bulb.
  • Working: light of suitable wavelength on the cathode ejects photoelectrons, the applied field drives them to the anode, and the microammeter reads the photocurrent.
  • Uses: control systems, light-measuring devices, light meters in photographic cameras, switching street lights on and off, counting devices, burglar alarms and traffic-defaulter alarms (an interrupted ultraviolet beam stops the photocurrent and triggers the counter or alarm).

14.5 De Broglie Hypothesis

  • De Broglie (1924) argued from the symmetry of nature: if radiation has a dual nature, matter should have one too.
  • For a photon, p=E/c=hν/c=h/λp = E/c = h\nu/c = h/\lambda. De Broglie proposed that a moving material particle of total energy EE and momentum pp has a wave associated with it in the same way, with ν=E/h\nu = E/h and λ=h/p=h/mv\lambda = h/p = h/mv.
  • These waves are matter waves and λ\lambda is the de Broglie wavelength. Greater the momentum, shorter the wavelength.
  • Useful forms: λ=h/2mEK\lambda = h/\sqrt{2mE_K}, and λ=h/2mqV\lambda = h/\sqrt{2mqV} for a charge qq accelerated from rest through VV volt. For an electron this reduces to λ=1.228/V\lambda = 1.228/\sqrt{V} nm.
  • The relation holds for electrons, protons and charged ions, but not when VV is so large that the speed approaches cc.
  • A student of 4545 kg running at 8 km h−18\ \text{km h}^{-1} has λ=6.63×10−36\lambda = 6.63\times10^{-36} m and a 12001200 kg car at 60 km h−160\ \text{km h}^{-1} has λ=3.32×10−38\lambda = 3.32\times10^{-38} m. These are negligible against the size of the object and the width of the path, which is why we never notice the wave nature of macroscopic bodies.
  • An electron of 100100 eV has λ=1.228\lambda = 1.228 angstrom, comparable to the size of an atom, so its wave nature is easy to see.

14.6 Davisson and Germer Experiment

  • Purpose: to test experimentally whether electrons, which are particles, show diffraction, which is a property of waves.
  • Set-up: an evacuated chamber containing an electron gun (tungsten filament F heated by battery B), a cylindrical anode that accelerates and collimates the beam, a nickel crystal target, a movable electron detector on a circular scale and a galvanometer.
  • The scattered intensity was not uniform in all directions as classical theory predicted; it showed peaks (constructive interference) and troughs (destructive interference), that is, a diffraction pattern.
  • Varying the accelerating potential from 4444 V to 6868 V, a sharp peak appeared at scattering angle 50∘50^\circ for V=54V = 54 V.
  • Theory: λ=1.228/54=0.167\lambda = 1.228/\sqrt{54} = 0.167 nm. Experiment gave 0.1650.165 nm. The close agreement substantiated de Broglie's hypothesis and verified his relation.
  • G. P. Thomson obtained the same conclusion in 1928 by diffracting electrons through metal foils. De Broglie got the Nobel Prize in 1929; Davisson, Germer and Thomson shared it in 1937. J. J. Thomson discovered the electron as a particle and his son G. P. Thomson showed it behaves as a wave.

14.7 Wave-Particle Duality of Matter

  • Material particles show wave-like nature under certain circumstances; this is the wave-particle duality of matter.
  • Waves are described classically by frequency and wave number, particles by mass and momentum. Duality means every moving particle has an associated frequency and wave number, and every wave has an associated energy and momentum.
  • Quantum behaviour becomes noticeable only for microscopic objects, of size about 10−610^{-6} m or less. The small magnitude of hh fixes the scale at which quantum phenomena appear.
  • Whenever the wavelength is small compared with the dimensions of slits or obstacles, or is too small to measure, Newtonian mechanics is enough.
  • In conclusion: for both electromagnetic radiation and atomic or sub-atomic particles, the particle nature dominates during interaction with matter, while the wave nature dominates during travel through space, particularly when the confinement is of the same order as the associated wavelength.
  • Three ways of supplying an electron the work function needed to escape: thermionic emission (heating to about 2000 ∘2000\ ^\circC), field emission (a strong field of about 106 V m−110^{6}\ \text{V m}^{-1} at a metal tip) and photoelectric emission (shining ultraviolet or visible radiation).
  • Electron microscope: resolution improves as the probing wavelength shortens. An electron beam of a few keV has a de Broglie wavelength far below an angstrom, so the resolution is several hundred times better than an optical microscope. Electrons also do not penetrate matter as light or X-rays do, are easily produced and controlled by electric and magnetic fields, and can be focused like light.

Distinctions the board asks for

Photo-emissive effectPhoto-voltaic effect
Electrons are ejected out of a photosensitive metal surfaceA potential difference is generated inside a semiconductor
The emitted electrons must be collected by an anode in an external circuitNo emission of electrons; the cell itself acts as a source of emf
Used in a photo cellUsed in a solar cell
Wave nature of radiationParticle nature of radiation
Shows in interference, diffraction and polarizationShows in black body radiation, photoelectric effect and Compton scattering
Described by frequency, wavelength and wave numberDescribed by photon energy hνh\nu and momentum h/λh/\lambda
Dominant while radiation travels through spaceDominant while radiation interacts with matter
Depends on intensityDepends on frequency
PhotocurrentMaximum kinetic energy of photoelectrons
Saturation currentStopping potential
Number of photoelectrons emitted per secondWhether emission happens at all

Standard graphs to be able to draw

  • Photocurrent against intensity, at fixed frequency and fixed accelerating potential: a straight line through the origin.
  • Photocurrent against accelerating potential, at fixed frequency and three different intensities: three curves rising and flattening to three different saturation currents, but all cutting the voltage axis at the same stopping potential −V0-V_0.
  • Photocurrent against accelerating potential, at fixed intensity and three different frequencies: all curves flatten to the same saturation current, but the stopping potentials −V01-V_0^1, −V02-V_0^2, −V03-V_0^3 are different, the largest belonging to the highest frequency.
  • Stopping potential against frequency, for emitters of different metals: parallel straight lines, common slope h/eh/e, different intercepts ν0\nu_0 on the frequency axis and −ϕ0/e-\phi_0/e on the V0V_0 axis.

Traps that cost marks

  • Writing K.E.max=hνK.E._{max} = h\nu and forgetting ϕ0\phi_0. The equation is K.E.max=hν−ϕ0K.E._{max} = h\nu - \phi_0.
  • Comparing an energy in eV with an energy in joule. Convert ϕ0\phi_0 to joule (multiply by 1.6×10−191.6\times10^{-19}) before subtracting it from hc/λhc/\lambda.
  • Substituting K.E.maxK.E._{max} in eV into vmax=2 K.E.max/mv_{max} = \sqrt{2\,K.E._{max}/m}. It must be in joule, or the speed comes out wrong by a factor of about 10910^{9} under the root.
  • Saying photoemission needs high intensity. It needs high enough frequency; intensity only controls how many electrons come out.
  • Saying stopping potential depends on intensity. It does not; it depends on the incident frequency and on the emitter material.
  • With polychromatic light, the stopping potential is set by the shortest wavelength (highest frequency) present, not the average.
  • Confusing the two threshold inequalities. Emission requires ν>ν0\nu > \nu_0 but λ<λ0\lambda < \lambda_0 — the wavelength condition reverses.
  • Using p=E/cp = E/c for an electron. That expression is only for a massless particle moving at cc, i.e. a photon. For an electron use p=mvp = mv or p=2mEKp = \sqrt{2mE_K}.
  • Forgetting that an electron at rest has p=0p = 0, so its de Broglie wavelength is infinite, i.e. undefined. Only a moving particle has a de Broglie wavelength.
  • Using λ=1.228/V\lambda = 1.228/\sqrt{V} and then reporting the answer in angstrom. That formula gives nanometre; the angstrom form is 12.28/V12.28/\sqrt{V}.
  • Applying λ=h/2mqV\lambda = h/\sqrt{2mqV} with VV in kilovolt without noting that the relation breaks down as the speed approaches cc.
  • Quoting the Davisson-Germer peak as being at 54∘54^\circ and 5050 V. It is at scattering angle 50∘50^\circ for an accelerating potential of 5454 V.
  • Saying red light can never cause photoemission. It can, if the metal has a low enough work function; the correct answer is that any radiation of sufficient energy will do.
  • Forgetting that the photocurrent stops the instant the light is switched off, since no photons remain to transfer energy.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 2 have come up in two or more papers.2
  1. 1
    de Broglie wavelength -- general relation Repeats
    What is de-Broglie's hypothesis? Obtain the relation for de-Broglie wavelength.

    Asked in: March 2022 (Q.31, 4 marks); July 2022 (Q.16, 3 marks)

    Show the derivation
    Step 1
    De Broglie proposed that a moving material particle of total energy E and momentum p has a wave associated with it (analogous to a photon).
    Step 2
    He suggested a relation between properties of wave, like frequency and wavelength, with that of a particle, like energy and momentum.
    p=Ec=hνc=hλp=\dfrac{E}{c}=\dfrac{h\nu}{c}=\dfrac{h}{\lambda}
    Step 3
    Thus frequency and wavelength of a wave associated with a material particle, of mass m moving with a velocity v, are given as
    ν=Eh and λ=hp=hmv....(1)\nu=\dfrac{E}{h}\text{ and }\lambda=\dfrac{h}{p}=\dfrac{h}{mv}\qquad....(1)
    Step 4
    De Broglie referred to these waves associated with material particles as matter waves. The wavelength of the matter waves, given by equation (1), is now known as de Broglie wavelength and the equation is known as de Broglie relation.
  2. 2
    de Broglie wavelength of an electron accelerated through potential difference V Repeats
    Obtain an expression for the de-Broglie wavelength associated with an electron accelerated from rest through a potential difference of 'V' volts.

    Asked in: October 2008; February 2026 (Q.20, 3 marks)

    Show the derivation
    Step 1
    Consider an electron of mass m and charge q is accelerated through potential difference of V volts.
    Step 2
    According to de Broglie hypothesis, wavelength of any particle is,
    λ=hp=hmv\lambda=\dfrac{h}{p}=\dfrac{h}{mv}
    ∵p=mv\because p=mv
    ∴m2v2=p2\therefore m^2v^2=p^2
    (12mv2)m=12p2\left(\dfrac{1}{2}mv^2\right)m=\dfrac{1}{2}p^2
    ∴E=p22m\therefore E=\dfrac{p^2}{2m}
    ∴p=2mE\therefore p=\sqrt{2mE}
    ∴λ=h2mE\therefore\lambda=\dfrac{h}{\sqrt{2mE}}
    Step 3
    Now, for an electron accelerated through potential difference V, the work done on the electron is stored in it in the form of energy.
    ∴E=qV\therefore E=qV
    ∴λ=h2mE=h2mqV\therefore\lambda=\dfrac{h}{\sqrt{2mE}}=\dfrac{h}{\sqrt{2mqV}}
    Step 4
    But, q=1.6×10−19q=1.6\times10^{-19} C, m=9.1×10−31m=9.1\times10^{-31} kg, h=6.63×10−34h=6.63\times10^{-34} J-s
    ∴λ=h2mqV\therefore\lambda=\dfrac{h}{\sqrt{2mqV}}
    =6.63×10−342×9.1×10−31×1.6×10−19×V(in volts)=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times V(\text{in volts})}}
    =1.228×10−9V(in volts) m=\dfrac{1.228\times10^{-9}}{\sqrt{V(\text{in volts})}}\text{ m}
    i.e., λ=1.228V(in volts) nm\lambda=\dfrac{1.228}{\sqrt{V(\text{in volts})}}\text{ nm}

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