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CalibratePhysics

Chapter 11

Dual Nature of Radiation and Matter

CBSE · 7 subtopics

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

Constants and conversions CBSE expects you to know

  • QuantityValue to quoteOther useful form
  • Planck's constant hh6.63×10−346.63\times10^{-34} J s4.14×10−154.14\times10^{-15} eV s
  • Speed of light cc3×1083\times10^{8} m s−1^{-1}c=νλc=\nu\lambda
  • Product hchc1.99×10−251.99\times10^{-25} J m12401240 eV nm, i.e. 1240012400 eV Å
  • Electron charge ee1.6×10−191.6\times10^{-19} C11 eV =1.6×10−19=1.6\times10^{-19} J
  • Electron mass mem_e9.1×10−319.1\times10^{-31} kg9.11×10−319.11\times10^{-31} kg
  • Specific charge e/me/m1.76×10111.76\times10^{11} C kg−1^{-1}same for all cathode materials
  • Proton mass mpm_p1.67×10−271.67\times10^{-27} kgmp≈1836 mem_p\approx1836\,m_e
  • hc=1240hc=1240 eV nm is the single most time-saving number in this chapter: photon energy in eV =1240/λ=1240/\lambda with λ\lambda in nm. Some CBSE keys use 12421242 eV nm (from h=4.14×10−15h=4.14\times10^{-15} eV s); either is accepted.
  • 11 Å =10−10=10^{-10} m =0.1=0.1 nm. So EE in eV =12400/λ=12400/\lambda with λ\lambda in Å.

Electron emission

FormulaSymbols, units and conditions
ϕ0=hν0=hcλ0\phi_0 = h\nu_0 = \dfrac{hc}{\lambda_0}
  • ϕ0\phi_0 (also written W0W_0) = work function, the minimum energy needed by an electron to escape the metal surface, in J or eV; ν0\nu_0 = threshold frequency (Hz); λ0\lambda_0 = threshold wavelength (m). Depends on the metal and on the state of its surface, not on the incident light.
λ0=cν0=hcϕ0\lambda_0 = \dfrac{c}{\nu_0} = \dfrac{hc}{\phi_0}
  • Threshold wavelength: with ϕ0\phi_0 in eV, λ0=1240/ϕ0\lambda_0 = 1240/\phi_0 nm. Emission needs λ≤λ0\lambda \le \lambda_0 (short wavelength), i.e. ν≥ν0\nu \ge \nu_0.
1 eV=1.6×10−19 J1\ \text{eV} = 1.6\times10^{-19}\ \text{J}
  • One electron volt is the energy gained by an electron accelerated through a potential difference of 11 volt.

Experimental study of the photoelectric effect

FormulaSymbols, units and conditions
Kmax=eV0=12mvmax2K_{max} = eV_0 = \dfrac12 m v_{max}^2
  • V0V_0 = stopping (cut-off) potential in volt, KmaxK_{max} = maximum kinetic energy of the photoelectrons, vmaxv_{max} = maximum speed. If V0V_0 is in volt, KmaxK_{max} in eV is numerically equal to V0V_0.
vmax=2eV0m=2Kmaxmv_{max} = \sqrt{\dfrac{2eV_0}{m}} = \sqrt{\dfrac{2K_{max}}{m}}
  • Maximum speed of the emitted photoelectrons, in m s−1^{-1}, with m=9.1×10−31m=9.1\times10^{-31} kg.
isat∝I(fixed ν>ν0)i_{sat} \propto I \qquad (\text{fixed } \nu > \nu_0)
  • Saturation photocurrent is directly proportional to the intensity II of the incident light, because the number of photoelectrons per second is proportional to the number of photons per second. V0V_0 does not change with II.

Einstein's photoelectric equation

FormulaSymbols, units and conditions
Kmax=hν−ϕ0orhν=12mvmax2+ϕ0K_{max} = h\nu - \phi_0 \qquad \text{or} \qquad h\nu = \dfrac12 m v_{max}^2 + \phi_0
  • The equation itself: photon energy hνh\nu = work function + maximum kinetic energy. Valid only for ν≥ν0\nu \ge \nu_0.
Kmax=h(ν−ν0)=hc(1λ−1λ0)K_{max} = h(\nu - \nu_0) = hc\left(\dfrac{1}{\lambda} - \dfrac{1}{\lambda_0}\right)
  • Same equation in frequency-difference and wavelength form. The wavelength form is the fastest route for numericals given λ\lambda and ϕ0\phi_0.
eV0=hν−ϕ0⟹V0=heν−ϕ0eeV_0 = h\nu - \phi_0 \qquad \Longrightarrow \qquad V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}
  • The stopping-potential line. Slope h/e=4.14×10−15h/e = 4.14\times10^{-15} V s, the same for every metal; intercept on the ν\nu-axis is ν0\nu_0; intercept on the V0V_0-axis is −ϕ0/e-\phi_0/e.
h=e ΔV0Δν=e×(slope of the V0–ν line)h = \dfrac{e\,\Delta V_0}{\Delta \nu} = e \times (\text{slope of the } V_0\text{--}\nu \text{ line})
  • How Millikan measured hh, and how CBSE asks you to read hh off a graph: take two points on the straight line and divide.
V0=hce⋅1λ−ϕ0e⟹h=e×(slope)cV_0 = \dfrac{hc}{e}\cdot\dfrac{1}{\lambda} - \dfrac{\phi_0}{e} \qquad \Longrightarrow \qquad h = \dfrac{e\times(\text{slope})}{c}
  • For a V0V_0 versus 1/λ1/\lambda graph, the slope is hc/ehc/e.
vmax2=2hcm⋅1λ−2ϕ0mv_{max}^2 = \dfrac{2hc}{m}\cdot\dfrac{1}{\lambda} - \dfrac{2\phi_0}{m}
  • For a vm2v_m^2 versus 1/λ1/\lambda graph: slope =2hcm=tan⁡θ=\dfrac{2hc}{m}=\tan\theta, so h=mtan⁡θ2ch=\dfrac{m\tan\theta}{2c} and ϕ0=mtan⁡θ2λ0\phi_0=\dfrac{m\tan\theta}{2\lambda_0}, where 1/λ01/\lambda_0 is the intercept on the 1/λ1/\lambda axis.
λ0=λ1λ22λ2−λ1ϕ0=hc(2λ2−λ1)λ1λ2\lambda_0 = \dfrac{\lambda_1\lambda_2}{2\lambda_2 - \lambda_1} \qquad \phi_0 = \dfrac{hc(2\lambda_2-\lambda_1)}{\lambda_1\lambda_2}
  • Standard result when KmaxK_{max} doubles as the wavelength changes from λ1\lambda_1 to λ2\lambda_2. Derive it in the exam; quote it only as a check.

Particle nature of light: the photon

FormulaSymbols, units and conditions
E=hν=hcλE = h\nu = \dfrac{hc}{\lambda}
  • Energy of one photon, in J (or eV if hh is in eV s). Independent of the intensity of the beam.
p=Ec=hνc=hλp = \dfrac{E}{c} = \dfrac{h\nu}{c} = \dfrac{h}{\lambda}
  • Momentum of a photon, in kg m s−1^{-1}. A photon has zero rest mass, travels at cc, is electrically neutral and is undeflected by electric and magnetic fields.
λ=hcE\lambda = \dfrac{hc}{E}
  • Wavelength of a photon of energy EE; with EE in eV, λ=1240/E\lambda = 1240/E nm.
N=Phν=PλhcN = \dfrac{P}{h\nu} = \dfrac{P\lambda}{hc}
  • NN = number of photons emitted per second by a source of power PP watt. The commonest one-step numerical in this chapter.
I=n hνA t=NA hνI = \dfrac{n\,h\nu}{A\,t} = N_A\,h\nu
  • Intensity in the photon picture: energy carried by the photons crossing unit area per unit time. nn = number of photons crossing area AA in time tt; NAN_A = number of photons per unit area per unit time. SI unit W m−2^{-2} (equivalently J m−2^{-2} s−1^{-1}).
νmax=eVhλmin=hceV\nu_{max} = \dfrac{eV}{h} \qquad \lambda_{min} = \dfrac{hc}{eV}
  • Cut-off of the X-ray continuous spectrum produced by electrons accelerated through VV volt (NCERT Exercise 11.1). With VV in volt, λmin=1240/V\lambda_{min} = 1240/V nm.

Wave nature of matter

FormulaSymbols, units and conditions
λ=hp=hmv\lambda = \dfrac{h}{p} = \dfrac{h}{mv}
  • de Broglie wavelength of a particle of mass mm (kg) moving with speed vv (m s−1^{-1}); λ\lambda in metre. Holds for every moving particle, charged or neutral, and is independent of the charge and the nature of the particle.
λ=h2mK\lambda = \dfrac{h}{\sqrt{2mK}}
  • In terms of kinetic energy KK (J), using p=2mKp=\sqrt{2mK}. Non-relativistic only. Gives λ∝K−1/2\lambda \propto K^{-1/2}.
λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}
  • For a particle of charge qq accelerated from rest through a potential difference VV volt, since K=qVK=qV. Gives λ∝V−1/2\lambda \propto V^{-1/2} and λ∝(mq)−1/2\lambda \propto (mq)^{-1/2}.
λ=12.27V A˚=1.227V nm\lambda = \dfrac{12.27}{\sqrt{V}}\ \text{\text{Å}} = \dfrac{1.227}{\sqrt{V}}\ \text{nm}
  • Electron accelerated through VV volt. Memorise this; it turns a two-line calculation into one. At V=100V=100 V, λ=1.227\lambda=1.227 Å.
p=2mK=2mqVp = \sqrt{2mK} = \sqrt{2mqV}
  • Momentum of an accelerated particle, in kg m s−1^{-1}; note p∝mp \propto \sqrt m at the same KK or same qVqV.
λphoton=hp=hhν/c=cν\lambda_{photon} = \dfrac{h}{p} = \dfrac{h}{h\nu/c} = \dfrac{c}{\nu}
  • Consistency check: the de Broglie wavelength of a photon equals the wavelength of the electromagnetic radiation it belongs to.
λ1λ2=m2q2m1q1(same V)λ1λ2=m2m1(same K)λ1λ2=m2m1(same v)\dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{m_2 q_2}{m_1 q_1}} \quad (\text{same } V) \qquad \dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{m_2}{m_1}} \quad (\text{same } K) \qquad \dfrac{\lambda_1}{\lambda_2} = \dfrac{m_2}{m_1} \quad (\text{same } v)
  • The three ratio formulas that cover every proton / deuteron / alpha-particle / electron comparison CBSE sets. Masses: md=2mpm_d = 2m_p, mα=4mpm_\alpha = 4m_p; charges: qp=qd=eq_p = q_d = e, qα=2eq_\alpha = 2e.
Resolving power∝1λ\text{Resolving power} \propto \dfrac{1}{\lambda}
  • Why an electron microscope beats an optical one: at 5050 kV the electron's de Broglie wavelength is about 5.5×10−125.5\times10^{-12} m, some 10510^{5} times smaller than yellow light, so the resolving power is about 10510^{5} times greater.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.7 subtopics
  1. 1Electron Emission1 board question
  2. 2Photoelectric Effect2 board questions
  3. 3Experimental Study of Photoelectric Effect20 board questions
  4. 4Photoelectric Effect and Wave Theory of Light1 board question
  5. 5Einstein's Photoelectric Equation: Energy Quantum of Radiation26 board questions
  6. 6Particle Nature of Light: The Photon5 board questions
  7. 7Wave Nature of Matter20 board questions
The whole chapter in brief

11.1 Introduction: how the electron was found

  • Discharge of electricity through gases at low pressure (about 0.0010.001 mm of mercury) produces cathode rays. Crookes (1870) discovered them; he suggested in 1879 that they are streams of fast negatively charged particles.
  • J. J. Thomson (1897) applied mutually perpendicular electric and magnetic fields across a discharge tube and measured the speed and the specific charge e/me/m of these particles. Speeds came out as 0.1c0.1c to 0.2c0.2c; the accepted value of e/me/m is 1.76×10111.76\times10^{11} C kg−1^{-1}.
  • e/me/m was found to be independent of the cathode material and of the gas in the tube — proof that these particles are universal constituents of matter. Thomson named them electrons and received the 1906 Nobel Prize.
  • Millikan's oil-drop experiment (1913) showed that the charge on a drop is always an integral multiple of 1.602×10−191.602\times10^{-19} C, so electric charge is quantised. Knowing ee and e/me/m gives the electron mass m=9.1×10−31m = 9.1\times10^{-31} kg.
  • Roentgen discovered X-rays in 1895. Particles emitted by UV-irradiated metals and by hot metals were found to have the same e/me/m as cathode rays — the same particle in every case.

Electron emission and work function

  • Metals have free electrons, but they cannot leave the surface: an escaping electron leaves the surface positively charged, and is pulled back.
  • Work function ϕ0\phi_0: the minimum energy required by an electron to escape from the metal surface. Measured in eV; 11 eV =1.6×10−19=1.6\times10^{-19} J. It depends on the metal and on the nature of its surface.
  • Three ways of supplying that energy:
ProcessEnergy sourceExample
Thermionic emissionheating the metalcathode of a vacuum tube
Field emissiona very strong field, about 10810^{8} V m−1^{-1}spark plug
Photoelectric emissionlight of suitable frequencyphotocell

Photoelectric effect: the discovery

  • Photoelectric effect: the emission of electrons from a metal surface when light of suitable frequency falls on it. The emitted electrons are called photoelectrons.
  • Hertz (1887): high-voltage sparks across his detector loop were enhanced when the emitter plate was illuminated by ultraviolet light from an arc lamp.
  • Hallwachs (1888): a negatively charged zinc plate connected to an electroscope lost its charge under UV light; an uncharged plate became positively charged; a positively charged plate became more positive. Conclusion: negatively charged particles are emitted.
  • Lenard (1886–1902): UV light on the emitter plate of an evacuated tube with two electrodes drove a current; the current stopped the instant the light was stopped.
  • Below a certain minimum frequency — the threshold frequency ν0\nu_0, characteristic of the emitter material — nothing is emitted, however intense the light.
  • Zinc, cadmium and magnesium respond only to ultraviolet light. The alkali metals lithium, sodium, potassium, caesium and rubidium are sensitive even to visible light.

The experimental arrangement

  • An evacuated glass or quartz tube contains a photosensitive plate C (the emitter/cathode) and a metal plate A (the collector/anode). The quartz window W lets ultraviolet light through.
  • Monochromatic light from source S passes through W onto C. Electrons emitted from C are collected by A.
  • A battery with a sliding contact sets the potential difference between C and A; a commutator reverses the polarity, so A can be made positive or negative with respect to C.
  • A voltmeter reads the plate potential; a microammeter reads the photocurrent.
  • The frequency is changed with coloured filters; the intensity is changed by moving the source nearer or further away.

The three graphs CBSE asks for

  • Photocurrent versus intensity (collector positive, frequency and potential fixed): a straight line through the origin. Photocurrent is directly proportional to intensity, because the number of photoelectrons per second is proportional to the number of photons per second.
  • Photocurrent versus collector plate potential: as the positive potential of A is raised, the current rises and then flattens at the saturation current, when every emitted electron reaches A. As A is made negative, the current falls and drops sharply to zero at the stopping potential −V0-V_0. For a fixed frequency and increasing intensities I1<I2<I3I_1 < I_2 < I_3: three curves of the same shape, all cutting the potential axis at the same −V0-V_0, saturating at successively higher currents. For a fixed intensity and increasing frequencies ν1<ν2<ν3\nu_1 < \nu_2 < \nu_3: three curves saturating at the same current, but cutting the potential axis at successively more negative stopping potentials.
  • Stopping potential versus frequency: a straight line, V0=(h/e)ν−ϕ0/eV_0 = (h/e)\nu - \phi_0/e. Slope h/eh/e, the same for every material; it meets the ν\nu-axis at ν0\nu_0 and the V0V_0-axis at −ϕ0/e-\phi_0/e. Two different metals give two parallel lines; the line further to the right has the greater ν0\nu_0, hence the greater work function.
  • Related straight lines: KmaxK_{max} versus ν\nu has slope hh and ν\nu-intercept ν0\nu_0; V0V_0 versus 1/λ1/\lambda has slope hc/ehc/e; vm2v_m^2 versus 1/λ1/\lambda has slope 2hc/m2hc/m.

The four experimental observations (learn this list)

  • For a given material and a frequency above threshold, the photocurrent is directly proportional to the intensity of the incident light.
  • For a given material and frequency, the saturation current is proportional to intensity, while the stopping potential is independent of intensity.
  • For a given material there is a threshold frequency ν0\nu_0 below which no emission occurs, however intense the light. Above ν0\nu_0, the stopping potential — equivalently KmaxK_{max} — increases linearly with frequency and is independent of intensity.
  • Photoelectric emission is instantaneous: the time lag is about 10−910^{-9} s or less, even for very dim light.

Definitions worth writing verbatim

  • Threshold frequency: the minimum frequency of incident radiation below which photoelectric emission does not take place, however large the intensity of light may be.
  • Stopping potential: the minimum negative (retarding) potential given to the collector plate with respect to the emitter at which the photoelectric current becomes zero.
  • Saturation current: the maximum photocurrent, reached when all the photoelectrons emitted per second are collected.
  • Intensity, in the photon picture: the light energy — the number of photons multiplied by the energy of each photon — incident per unit area per unit time. SI unit W m−2^{-2}.
  • Work function: the minimum energy required by an electron to escape from the metal surface.

Why the wave theory fails

Wave theory predictsExperiment shows
KmaxK_{max} should rise with intensity, since the field amplitude risesKmaxK_{max} is completely independent of intensity
Any frequency should work if the beam is intense enough — no thresholdNothing is emitted below ν0\nu_0, however intense the beam
Energy is spread over the wavefront, so an electron must accumulate it over hoursEmission is instantaneous, within 10−910^{-9} s
  • The stopping potential — independent of intensity, dependent on frequency — is the crucial discriminator between the wave picture and the photon picture.

Einstein's photoelectric equation (1905)

  • Radiation energy comes in discrete quanta, each of energy hνh\nu. In photoemission a single electron absorbs a single quantum, all of it, at once.

Kmax=hν−ϕ0K_{max} = h\nu - \phi_0

  • It explains all four observations at once: KmaxK_{max} is linear in ν\nu and free of intensity; emission needs hν>ϕ0h\nu > \phi_0, i.e. ν>ν0=ϕ0/h\nu > \nu_0 = \phi_0/h; intensity fixes the number of quanta and hence the current; and one-photon-one-electron absorption is instantaneous.
  • Electrons deeper inside the metal are more tightly bound and emerge with less than KmaxK_{max}, so photoelectrons come out with a spread of energies from 00 to KmaxK_{max}.
  • Millikan set out (1906–1916) to disprove the equation, measured the slope of the V0V_0–ν\nu line for sodium, obtained h=6.626×10−34h = 6.626\times10^{-34} J s in agreement with Planck's value, and ended up proving it.

The photon picture: five properties

  • In its interaction with matter, radiation behaves as if made of particles called photons.
  • Each photon has energy E=hνE = h\nu, momentum p=hν/c=h/λp = h\nu/c = h/\lambda, and speed cc.
  • All photons of a given frequency carry the same energy and momentum whatever the intensity. Raising the intensity increases only the number of photons crossing unit area per second.
  • Photons are electrically neutral and are not deflected by electric or magnetic fields.
  • In a photon-particle collision, total energy and total momentum are conserved, but the number of photons need not be — a photon may be absorbed or created.
  • Compton's 1924 experiment on the scattering of X-rays by electrons confirmed the particle behaviour of light. Einstein received the 1921 Nobel Prize (partly for the photoelectric effect) and Millikan the 1923 Nobel Prize.

Wave nature of matter: de Broglie

  • de Broglie's argument (1924): nature is symmetrical, and matter and radiation are the two basic entities; since radiation shows a dual character, matter should too. So a wave should be associated with every moving material particle.

λ=hp=hmv\lambda = \dfrac{h}{p} = \dfrac{h}{mv}

  • The left side is a wave attribute, the right side a particle attribute, and Planck's constant links them — the dual nature of matter is contained in this one relation.
  • The de Broglie wavelength is independent of the charge and the nature of the particle; it depends only on the momentum.
  • Consistency with radiation: for a photon p=hν/cp = h\nu/c, so λ=h/p=c/ν\lambda = h/p = c/\nu, exactly the wavelength of the radiation.
  • Why we do not see matter waves: a 0.120.12 kg ball at 2020 m s−1^{-1} has p=2.40p = 2.40 kg m s−1^{-1} and λ=2.76×10−34\lambda = 2.76\times10^{-34} m — far beyond measurement. An electron at 5.4×1065.4\times10^{6} m s−1^{-1} has λ=0.135\lambda = 0.135 nm, comparable to X-ray wavelengths and to the spacing of atomic planes in a crystal, so it is measurable.
  • The phase velocity of a matter wave has no physical significance; the group velocity does, and equals the velocity of the particle.

Standard ratio results

SituationResult
Same speed vvλ∝1/m\lambda \propto 1/m
Same kinetic energy KKλ∝1/m\lambda \propto 1/\sqrt m
Same accelerating potential VVλ∝1/mq\lambda \propto 1/\sqrt{mq}
Proton and α\alpha-particle, same VVλp:λα=22:1\lambda_p : \lambda_\alpha = 2\sqrt2 : 1
Deuteron and α\alpha-particle, same VVλd:λα=2:1\lambda_d : \lambda_\alpha = 2 : 1
Proton and α\alpha-particle, same KKλp:λα=2:1\lambda_p : \lambda_\alpha = 2 : 1
Proton and deuteron, same speedλp:λd=2:1\lambda_p : \lambda_d = 2 : 1
Proton and electron, same speedλp:λe=1:1836\lambda_p : \lambda_e = 1 : 1836
Two fragments of a body at restλ1:λ2=1:1\lambda_1 : \lambda_2 = 1 : 1 (equal and opposite momenta)

Numbers CBSE reuses

MetalWork function (eV)Threshold wavelength (nm)
Caesium Cs2.142.14580580
Potassium K2.32.3539539
Sodium Na2.752.75451451
Calcium Ca3.203.20388388
Molybdenum Mo4.174.17297297
Nickel Ni5.155.15241241
  • Board papers have also quoted Na as 1.921.92 eV and K as 2.152.15 eV in one 2018 question; always use the value printed in the paper, not a remembered one.

Traps that cost marks

  • Stopping potential depends on frequency and on the material only. Doubling the intensity, or moving the lamp closer, changes nothing about V0V_0 or KmaxK_{max} — only the current changes. This single point is examined nearly every year.
  • Moving a source closer or further changes intensity, not frequency. A metal that would not emit still will not emit.
  • Threshold is a condition on frequency, so on wavelength it reverses: emission needs ν≥ν0\nu \ge \nu_0 but λ≤λ0\lambda \le \lambda_0. A longer wavelength is a lower frequency.
  • "Greater threshold frequency" means "greater work function" and "smaller threshold wavelength". Keep the three straight.
  • The slope of the V0V_0–ν\nu line is h/eh/e and is the same for all metals. Only the intercept changes with the material. Do not say the slope depends on the metal.
  • Do not write "the kinetic energy of photoelectrons is hν−ϕ0h\nu-\phi_0". It is the maximum kinetic energy; electrons emerge with a spread of energies from zero up to that value, because they are not all equally bound.
  • Photocurrent is proportional to intensity, not to frequency; KmaxK_{max} is proportional to frequency (linearly, with an offset), not to intensity. Swapping these is the classic lost mark.
  • Kmax=hν−ϕ0K_{max} = h\nu - \phi_0 is linear in ν\nu but not proportional to it: doubling ν\nu gives 2K+ϕ02K+\phi_0, which is more than 2K2K.
  • eV0eV_0 in joule equals KmaxK_{max} in joule; if V0V_0 is in volt then KmaxK_{max} in eV is the same number. Do not multiply by ee twice.
  • Work function must be converted to joule before dividing by hh, or else use h=4.14×10−15h = 4.14\times10^{-15} eV s throughout. Mixing units here wrecks the answer.
  • de Broglie wavelength applies to every moving particle, not only to charged ones, and does not depend on charge. But λ=h/2mqV\lambda = h/\sqrt{2mqV} does, because that route uses K=qVK = qV.
  • Do not use λ=h/2mqV\lambda = h/\sqrt{2mqV} for a neutral particle or for a particle whose kinetic energy was not acquired from an accelerating potential.
  • λ=12.27/V\lambda = 12.27/\sqrt V Å is for an electron only. Do not apply it to protons or alpha particles.
  • A photon has zero rest mass but non-zero momentum h/λh/\lambda; never write p=mvp = mv for a photon.
  • Photons are neutral, so they are not deflected by electric or magnetic fields — a favourite one-mark statement.
  • On a pp versus λ\lambda graph the curve is a rectangular hyperbola, not a straight line, since p=h/λp = h/\lambda.
  • In the photocurrent versus potential graph, remember the current is not zero at zero potential: some electrons reach the collector on their own kinetic energy. The curve meets the axis at the negative stopping potential.
  • State intensity in the photon picture as energy per unit area per unit time, and give the unit W m−2^{-2} when asked for it.
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.7
  1. 1
    Derivation 1
    The figure shows vm2v_m^2 versus 1/λ1/\lambda graph for photoelectrons emitted from a surface, where vmv_m is the maximum speed of electrons and λ\lambda is the wavelength of incident radiation (a straight line making angle θ\theta with the 1/λ1/\lambda axis, with a positive intercept on the 1/λ1/\lambda axis). Using this graph and Einstein's photoelectric equation, obtain the expression for Planck's constant and the work function of the surface.

    Asked in: CBSE Class 12 Physics — SA · 2023 · Section B, 2 marks

    Show the derivation

    Einstein's equation:   12mvm2=hν−ϕ0=hcλ−ϕ0\;\dfrac12mv_m^2=h\nu-\phi_0=\dfrac{hc}\lambda-\phi_0   vm2=2hcm⋅1λ−2ϕ0m\;v_m^2=\dfrac{2hc}m\cdot\dfrac1\lambda-\dfrac{2\phi_0}m This is linear in 1/λ1/\lambda with slope 2hcm=tan⁡θ\dfrac{2hc}m=\tan\theta, giving   h=mtan⁡θ2c\;h=\dfrac{m\tan\theta}{2c}. The line intersects the 1/λ1/\lambda axis (where vm2=0v_m^2=0) at 1/λ01/\lambda_0, the threshold-wavelength reciprocal, where ϕ0=hcλ0\phi_0=\dfrac{hc}{\lambda_0}. Substituting the value of hh found above:   ϕ0=mtan⁡θ2c×cλ0=mtan⁡θ2λ0\;\phi_0=\dfrac{m\tan\theta}{2c}\times\dfrac c{\lambda_0}=\dfrac{m\tan\theta}{2\lambda_0}, where 1/λ01/\lambda_0 is the graph's intercept on the 1/λ1/\lambda axis.

  2. 2
    Derivation 2
    The maximum kinetic energy of the photoelectrons emitted is doubled when the wavelength of light incident on the photosensitive surface changes from λ1\lambda_1 to λ2\lambda_2. Deduce expressions for the threshold wavelength and work function of the metal surface in terms of λ1\lambda_1 and λ2\lambda_2.

    Asked in: CBSE Class 12 Physics — SA · 2020 · Section C, 3 marks

    Show the derivation

      hcλ=Kmax+ϕ\;\dfrac{hc}\lambda=K_{max}+\phi For case I:   hcλ1=K+hcλT  ⇒  hc(1λ1−1λT)=K…(i)\;\dfrac{hc}{\lambda_1}=K+\dfrac{hc}{\lambda_T}\;\Rightarrow\;hc\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_T}\right)=K\quad\dots(i) For case II:   hcλ2=2K+hcλT  ⇒  hc(1λ2−1λT)=2K…(ii)\;\dfrac{hc}{\lambda_2}=2K+\dfrac{hc}{\lambda_T}\;\Rightarrow\;hc\left(\dfrac1{\lambda_2}-\dfrac1{\lambda_T}\right)=2K\quad\dots(ii) Substituting (i) into (ii):   hc(1λ2−1λT)=2hc(1λ1−1λT)\;hc\left(\dfrac1{\lambda_2}-\dfrac1{\lambda_T}\right)=2hc\left(\dfrac1{\lambda_1}-\dfrac1{\lambda_T}\right)   1λ2−1λT=2λ1−2λT  ⇒  1λT=2λ1−1λ2  ⇒  λT=λ1λ22λ2−λ1\;\dfrac1{\lambda_2}-\dfrac1{\lambda_T}=\dfrac2{\lambda_1}-\dfrac2{\lambda_T}\;\Rightarrow\;\dfrac1{\lambda_T}=\dfrac2{\lambda_1}-\dfrac1{\lambda_2}\;\Rightarrow\;\lambda_T=\dfrac{\lambda_1\lambda_2}{2\lambda_2-\lambda_1} Work function,   W0=hcλT=hc(2λ2−λ1)λ1λ2\;W_0=\dfrac{hc}{\lambda_T}=\dfrac{hc(2\lambda_2-\lambda_1)}{\lambda_1\lambda_2}

  3. 3
    Derivation 3
    Write Einstein's photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation. The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from λ1\lambda_1 to λ2\lambda_2. Derive the expressions for the threshold wavelength λ0\lambda_0 and work function for the metal surface.

    Asked in: CBSE Class 12 Physics — SA · Delhi 2015, Cr · Section C, 3 marks

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    Einstein's photoelectric equation:   Kmax=hν−ϕ0=hcλ−ϕ0\;K_{max}=h\nu-\phi_0=\dfrac{hc}\lambda-\phi_0 Features explained: existence of threshold frequency (no emission below it, regardless of intensity), instantaneous emission, and KmaxK_{max} depending only on frequency (not intensity) while photocurrent depends only on intensity (not frequency, above threshold). Same numerical derivation as Q40/Q49 pattern (Q40 above): for wavelength λ1\lambda_1,   hcλ1=K+hcλ0\;\dfrac{hc}{\lambda_1}=K+\dfrac{hc}{\lambda_0}; for λ2\lambda_2 (giving 2K2K),   hcλ2=2K+hcλ0\;\dfrac{hc}{\lambda_2}=2K+\dfrac{hc}{\lambda_0}. Eliminating KK:   2(hcλ1−hcλ0)=hcλ2−hcλ0\;2\left(\dfrac{hc}{\lambda_1}-\dfrac{hc}{\lambda_0}\right)=\dfrac{hc}{\lambda_2}-\dfrac{hc}{\lambda_0}   ϕ0=2hcλ1−hcλ2=hc(2λ1−1λ2)\;\phi_0=\dfrac{2hc}{\lambda_1}-\dfrac{hc}{\lambda_2}=hc\left(\dfrac2{\lambda_1}-\dfrac1{\lambda_2}\right) Also   ϕ0=hcλ0  ∴  hcλ0=hc(2λ1−1λ2)  ⇒  1λ0=2λ2−λ1λ1λ2  ⇒  λ0=λ1λ22λ2−λ1\;\phi_0=\dfrac{hc}{\lambda_0}\;\therefore\;\dfrac{hc}{\lambda_0}=hc\left(\dfrac2{\lambda_1}-\dfrac1{\lambda_2}\right)\;\Rightarrow\;\dfrac1{\lambda_0}=\dfrac{2\lambda_2-\lambda_1}{\lambda_1\lambda_2}\;\Rightarrow\;\lambda_0=\dfrac{\lambda_1\lambda_2}{2\lambda_2-\lambda_1} Work function:   W0=hc(2λ2−λ1)λ1λ2\;W_0=\dfrac{hc(2\lambda_2-\lambda_1)}{\lambda_1\lambda_2}

  4. 4
    Derivation 4
    (a) Write the important properties of photons which are used to establish Einstein's photoelectric equation. (b) Use this equation to explain the concept of (i) threshold frequency and (ii) stopping potential.

    Asked in: CBSE Class 12 Physics — SA · AI 2015, U · Section C, 3 marks

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    (a) (i) In the interaction of photons with free electrons, the entire energy of the photon is absorbed. (ii) Energy of a photon is directly proportional to frequency; intensity of incident radiation depends on the number of photons falling per unit area per unit time for a given frequency. (iii) In a photon-electron collision, total energy and total momentum remain conserved. Einstein's equation: Kmax=hν−ϕ0K_{max}=h\nu-\phi_0. (b) (i) Einstein's photoelectric equation: according to Einstein, when light is incident on a metal surface, incident photons are absorbed completely by valence electrons of atoms on its surface. Energy hνh\nu of each photon is partially utilized by an electron to become free of its "work function" W0W_0, and the rest provides the maximum kinetic energy to the photoelectron, i.e. hν=12mvmax2+W0h\nu=\dfrac12mv_{max}^2+W_0. The minimum value of the frequency of incident radiation below which photoelectric emission stops (i.e. kinetic energy of photoelectron is zero) is called threshold frequency, ν0=W0/h\nu_0=W_0/h. (ii) When the work done by the collecting-electrode potential on a photoelectron equals its maximum kinetic energy, that electrode potential is known as the stopping potential, V0=Kmax/eV_0=K_{max}/e.

  5. 5
    Derivation 5
    The wavelength λ\lambda of a photon and the de Broglie wavelength of an electron have the same value. Show that the energy of a photon is (2λmc/h)(2\lambda mc/h) times the kinetic energy of the electron, where mm, cc and hh have their usual meaning.

    Asked in: CBSE Class 12 Physics — SA · Foreign 2016, An · Section B, 2 marks

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    de Broglie wavelength of electron:   λ=hmv  ⇒  v=hmλ\;\lambda=\dfrac h{mv}\;\Rightarrow\;v=\dfrac h{m\lambda} Kinetic energy of electron:   Ee=12mv2=12m(hmλ)2=h22mλ2…(i)\;E_e=\dfrac12mv^2=\dfrac12m\left(\dfrac h{m\lambda}\right)^2=\dfrac{h^2}{2m\lambda^2}\quad\dots(i) Energy of photon of the same wavelength λ\lambda:   Ep=hcλ…(ii)\;E_p=\dfrac{hc}\lambda\quad\dots(ii) Dividing (ii) by (i):   EpEe=hcλ×2mλ2h2=2mλch\;\dfrac{E_p}{E_e}=\dfrac{hc}\lambda\times\dfrac{2m\lambda^2}{h^2}=\dfrac{2m\lambda c}h ∴  Ep=2λmchEe\therefore\;E_p=\dfrac{2\lambda mc}hE_e, as required.

  6. 6
    Derivation 6
    Obtain an expression for the ratio of the accelerating potentials required to accelerate a proton and an α\alpha-particle to have the same de-Broglie wavelength associated with them.

    Asked in: CBSE Class 12 Physics — SA · AI 2019 · Section C, 3 marks

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      λ=h2mqV\;\lambda=\dfrac h{\sqrt{2mqV}}. Let proton (mp,qp=em_p,q_p=e) and α\alpha-particle (mα=4mp,qα=2em_\alpha=4m_p,q_\alpha=2e) have the same λ\lambda.   h2mpqpVp=h2mαqαVα  ⇒  mpqpVp=mαqαVα\;\dfrac h{\sqrt{2m_pq_pV_p}}=\dfrac h{\sqrt{2m_\alpha q_\alpha V_\alpha}}\;\Rightarrow\;m_pq_pV_p=m_\alpha q_\alpha V_\alpha   mp×e×Vp=4mp×2e×Vα  ⇒  Vp=8Vα  ⇒  VpVα=8:1\;m_p\times e\times V_p=4m_p\times2e\times V_\alpha\;\Rightarrow\;V_p=8V_\alpha\;\Rightarrow\;\dfrac{V_p}{V_\alpha}=8:1

  7. 7
    Derivation 7
    (a) Explain the de-Broglie argument to propose his hypothesis. Show that de-Broglie wavelength of a photon equals the wavelength of electromagnetic radiation. (b) If deuterons and an alpha particle are accelerated through the same potential, find the ratio of the associated de-Broglie wavelengths of the two.

    Asked in: CBSE Class 12 Physics — SA · 2020-21, An · Section C, 3 marks · CBSE sample question

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    (a) De Broglie reasoned that nature is symmetrical, and that the two basic physical entities of the universe — matter and radiation — should therefore also be symmetrical in nature. If radiation exhibits a dual (wave and particle) character, matter should too. This led him to propose that a wave of wavelength λ=h/p\lambda=h/p is associated with every moving material particle, called the de Broglie or matter wave. For a photon: momentum p=hνcp=\dfrac{h\nu}c (from E=hν=pcE=h\nu=pc). Substituting into the de Broglie relation,   λ=hp=hhν/c=cν\;\lambda=\dfrac hp=\dfrac h{h\nu/c}=\dfrac c\nu, which is exactly the wavelength of the electromagnetic radiation of frequency ν\nu — confirming consistency between the two pictures. (b)   λ=h2mqV\;\lambda=\dfrac h{\sqrt{2mqV}}. Deuteron: md, qd=em_d,\,q_d=e; alpha particle: mα=2mdm_\alpha=2m_d, qα=2eq_\alpha=2e (since mα≈4mp=2mdm_\alpha\approx4m_p=2m_d and qα=2e=2qdq_\alpha=2e=2q_d).   λdλα=mαqαmdqd=2md×2emd×e=4=2  ⇒  λd:λα=2:1\;\dfrac{\lambda_d}{\lambda_\alpha}=\sqrt{\dfrac{m_\alpha q_\alpha}{m_dq_d}}=\sqrt{\dfrac{2m_d\times2e}{m_d\times e}}=\sqrt4=2\;\Rightarrow\;\lambda_d:\lambda_\alpha=2:1

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