Hello dear learner. Welcome to HSLC GURU. This English-medium lesson presents the complete question-answer set, key concepts, derivations, formula sheet, illustrative diagrams and additional important questions of the eleventh chapter — Dual Nature of Radiation and Matter — of the ASSEB (Assam State School Education Board) Class 12 Physics syllabus. The chapter studies one of the most revolutionary discoveries of the twentieth century: that light is not purely a wave but also a stream of energy quanta called photons, and that material particles such as electrons are not purely tiny “billiard-balls” but also possess a wave character described by a de Broglie wavelength. The mutual interplay of these two natures forms the foundation of quantum mechanics and ushers in the modern study of atoms, nuclei and solids.

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Summary

Electrons are emitted from a metal surface only when the energy supplied to them exceeds a minimum value $\phi_0$ called the work function of the metal. The energy can be supplied as heat (thermionic emission), as light (photoelectric emission), as a strong electric field (field emission) or by impact of fast charged particles (secondary emission). When monochromatic light of frequency $\nu$ is incident on a clean metal surface, electrons are ejected promptly provided $\nu \ge \nu_0$, the threshold frequency. Below $\nu_0$ no current flows however intense the beam may be — a behaviour that classical wave theory cannot explain.

Albert Einstein (1905) resolved the puzzle by extending Max Planck’s quantum hypothesis: light of frequency $\nu$ travels as discrete quanta of energy $E=h\nu$ called photons. A photon either is absorbed completely by a single electron or not at all. Energy conservation then yields Einstein’s photoelectric equation $h\nu = \phi_0 + K_{\max}$. Stopping potential $V_0$ is related to maximum kinetic energy by $eV_0 = K_{\max}$, so $V_0 = (h/e)\nu – \phi_0/e$ — a straight line whose slope $h/e$ is independent of the metal and whose x-intercept is the threshold frequency. The Davisson–Germer experiment (1927) demonstrated electron diffraction from a nickel crystal, confirming Louis de Broglie’s hypothesis (1924) that every moving particle has a wavelength $\lambda = h/p = h/mv$. For an electron accelerated through a potential difference $V$ volts, $\lambda = 12.27/\sqrt{V}$ Å. Heisenberg’s uncertainty principle, $\Delta x \cdot \Delta p \ge h/(4\pi)$, expresses the inevitable spread that must accompany any matter wave. Together, photons and matter waves embody the wave-particle duality that is the cornerstone of modern physics.

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Key Formulas

Quantity	Formula	SI Unit
Energy of a photon	$E = h\nu = \dfrac{hc}{\lambda}$	joule (J)
Momentum of a photon	$p = \dfrac{h}{\lambda} = \dfrac{E}{c} = \dfrac{h\nu}{c}$	kg m s$^{-1}$
Mass-equivalent of a photon	$m = \dfrac{E}{c^2} = \dfrac{h\nu}{c^2}$	kilogram (kg)
Work function	$\phi_0 = h\nu_0 = \dfrac{hc}{\lambda_0}$	joule (J) or eV
Threshold frequency	$\nu_0 = \dfrac{\phi_0}{h}$	hertz (Hz)
Threshold wavelength	$\lambda_0 = \dfrac{hc}{\phi_0}$	metre (m)
Einstein’s photoelectric equation	$h\nu = \phi_0 + K_{\max}$	joule (J)
Maximum KE of photoelectron	$K_{\max} = \tfrac{1}{2}mv_{\max}^2 = h(\nu-\nu_0)$	joule (J)
Stopping potential	$eV_0 = K_{\max} \Rightarrow V_0 = \dfrac{h}{e}\nu – \dfrac{\phi_0}{e}$	volt (V)
de Broglie wavelength	$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$	metre (m)
de Broglie wavelength (KE)	$\lambda = \dfrac{h}{\sqrt{2mK}}$	metre (m)
Electron de Broglie wavelength	$\lambda = \dfrac{12.27}{\sqrt{V}}$ Å (V in volt)	ångström
Heisenberg uncertainty	$\Delta x \cdot \Delta p \ge \dfrac{h}{4\pi}$	J s
Number of photons per sec from a source of power $P$	$n = \dfrac{P}{h\nu} = \dfrac{P\lambda}{hc}$	s$^{-1}$
Planck’s constant	$h = 6.626\times10^{-34}$ J s	J s
Charge / mass of electron	$e=1.6\times10^{-19}$ C, $m_e=9.11\times10^{-31}$ kg	C, kg

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11.1 Electron Emission

Conduction electrons are not free to leave a metal at ordinary temperatures because of the attractive pull of the positive ions of the lattice. To escape, an electron must possess energy at least equal to the work function $\phi_0$ — the minimum energy required to liberate an electron from the metal surface. Work function is conventionally measured in electron-volt (eV); for example, caesium has $\phi_0 \approx 2.14$ eV while platinum has $\phi_0 \approx 5.65$ eV. The energy needed for emission can be supplied in four standard ways.

• Thermionic emission: the metal is heated so that thermal kinetic energy of conduction electrons exceeds $\phi_0$. The hot tungsten filament of a CRT or vacuum diode is the classical example.
• Photoelectric emission: photons of frequency $\nu \ge \nu_0$ deliver energy $h\nu$ to a surface electron, which then escapes with maximum kinetic energy $K_{\max}=h\nu-\phi_0$.
• Field (or cold-cathode) emission: a very strong external electric field ($\sim 10^8$ V m$^{-1}$) reduces the surface barrier so that electrons tunnel out without thermal assistance — used in scanning tunnelling microscopes and field-emission displays.
• Secondary emission: energetic electrons or ions striking a surface knock out further electrons. Photomultiplier tubes use cascaded secondary emission to amplify weak photon signals.

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11.2 Photoelectric Effect — Hertz, Hallwachs and Lenard

Heinrich Hertz (1887) while working with electromagnetic spark discharges noticed that the spark across the secondary gap was easier when the negative electrode was illuminated by ultraviolet light. He had stumbled upon the photoelectric effect — yet he did not pursue it.

Wilhelm Hallwachs (1888) repeated the experiment with a clean zinc plate connected to an electroscope. He found that an uncharged plate became positively charged, a positively charged plate became more positive, but a negatively charged plate lost its charge — all when illuminated by ultraviolet light. He concluded that light ejects negatively charged particles from the surface.

Philipp Lenard (1900) performed the first quantitative study using an evacuated glass tube with a metal cathode and a collector. Light incident on the cathode produced a measurable photoelectric current that depended on light intensity and the applied voltage. Lenard established three central observations: (i) the current is proportional to incident intensity; (ii) the kinetic energy of photoelectrons increases with the frequency of the incident light, but is independent of intensity; (iii) emission is essentially instantaneous — the time-lag is less than $10^{-9}$ s.

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11.3 Experimental Observations

(a) Effect of Intensity

Keeping the frequency $\nu \;(>\nu_0)$ and the accelerating potential $V$ fixed, the photoelectric current is found to be directly proportional to the intensity of the incident light. Doubling the intensity doubles the number of photoelectrons emitted per second, but does not change the maximum kinetic energy of any individual electron.

(b) Effect of Potential

For a given frequency and intensity, the photoelectric current rises with the positive collector potential and saturates at a value $I_s$. When the collector is made progressively negative (retarding potential), the current falls; at a particular negative potential $V_0$, called the stopping (or cut-off) potential, the current becomes zero. At this point the most energetic electron is just unable to reach the collector:

$$eV_0 = K_{\max} = \tfrac{1}{2}m v_{\max}^2$$

$V_0$ is independent of intensity but increases linearly with frequency.

(c) Effect of Frequency

For a fixed metal cathode, photoelectric emission occurs only if $\nu \ge \nu_0$. Below $\nu_0$ no current flows, regardless of intensity. Above $\nu_0$ the stopping potential varies linearly with $\nu$:

$$V_0 = \dfrac{h}{e}\nu – \dfrac{\phi_0}{e}$$

The slope of the $V_0$–$\nu$ graph equals $h/e$ and is the same for every photoemitter, while the $\nu$-axis intercept equals the threshold frequency $\nu_0$ characteristic of that metal.

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11.4 Failure of the Classical Wave Theory

According to classical electromagnetism, light is a continuous wave whose energy is proportional to the square of its amplitude (i.e. its intensity). Three predictions follow that completely contradict the observations:

1. Energy delivered to a single electron should depend on the intensity of light, not its frequency. Hence increasing intensity should increase the maximum kinetic energy of photoelectrons. Observation: $K_{\max}$ is independent of intensity.
2. For sufficiently weak light the electron should require a measurable time (often minutes or hours) to accumulate the threshold energy from a continuous wave. Observation: emission is prompt — within $10^{-9}$ s — even for very weak light.
3. Light of any frequency, however low, should liberate electrons provided the intensity is sufficiently large. Observation: no emission below $\nu_0$, irrespective of how intense the beam is.

The wave model therefore fails to explain (i) the existence of a threshold frequency, (ii) the instantaneous nature of emission, and (iii) the dependence of $K_{\max}$ on frequency rather than intensity.

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11.5 Einstein’s Photoelectric Equation

Einstein assumed that radiation of frequency $\nu$ is emitted, propagated and absorbed in indivisible packets of energy $E = h\nu$. When such a quantum is absorbed by an electron in the metal, the energy is used in two parts: $\phi_0$ to overcome the surface barrier and the remainder appears as the kinetic energy of the freed electron. Hence:

$$h\nu = \phi_0 + K_{\max}, \qquad K_{\max} = \tfrac{1}{2}m v_{\max}^2 = h\nu-\phi_0 = h(\nu-\nu_0)$$

Since $eV_0 = K_{\max}$, dividing throughout by $e$ gives the linear relation

$$V_0 = \dfrac{h}{e}\nu \;-\; \dfrac{\phi_0}{e}$$

This single equation explains every observed feature: the existence of a threshold ($K_{\max}\ge 0 \Rightarrow \nu\ge\nu_0$), the linear $V_0$–$\nu$ graph, the universality of slope $h/e$, the proportionality of current to intensity (more photons → more electrons) and the instantaneous nature of emission (one-photon-one-electron interaction). Robert Millikan (1916), although initially sceptical, performed precise measurements that confirmed Einstein’s formula and yielded a value of Planck’s constant in excellent agreement with that obtained from black-body radiation. Einstein received the 1921 Nobel Prize for this work.

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11.6 The Photon

The quantum of light is called a photon. Its principal properties are summarised below.

• Energy: $E = h\nu = hc/\lambda$.
• Speed: $c = 3\times 10^8$ m s$^{-1}$ in vacuum, irrespective of the source.
• Rest mass: zero. A photon exists only when in motion.
• Linear momentum: $p = E/c = h\nu/c = h/\lambda$.
• Equivalent (relativistic) mass: $m = E/c^2 = h\nu/c^2$.
• Photons are electrically neutral and are not deflected by electric or magnetic fields.
• Photon-electron collisions obey the conservation of energy and momentum (Compton scattering).
• The number of photons in a beam may change in interactions, but their individual energy depends only on $\nu$.

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11.7 Wave Nature of Matter — de Broglie Hypothesis

By 1923 the dual character of light — particle ($p=h/\lambda$) and wave ($\lambda$) — was firmly established. Louis de Broglie argued by symmetry that nature should be reciprocal: if waves can behave as particles, then particles should also behave as waves. He proposed that with every moving material particle of momentum $p=mv$ is associated a wavelength

$$\lambda \;=\; \dfrac{h}{p} \;=\; \dfrac{h}{mv}$$

For a particle of mass $m$ and kinetic energy $K = p^2/2m$, the wavelength may also be written as $\lambda = h/\sqrt{2mK}$. For an electron accelerated from rest through a potential difference of $V$ volts, $K = eV$ and substitution of numerical values gives the convenient form

$$\lambda \;=\; \dfrac{12.27}{\sqrt{V}}\;\text{Å}\quad(V\text{ in volt})$$

For a 100 V electron $\lambda \approx 1.23$ Å — comparable to inter-atomic spacings in crystals, so electrons can be diffracted by crystal lattices in the same way as X-rays. By contrast, a 60 g cricket ball moving at 30 m s$^{-1}$ has $\lambda \approx 3.7\times 10^{-34}$ m, so its wave nature is utterly unobservable. The de Broglie wave is not a material wave; it is a probability wave whose square gives the likelihood of finding the particle.

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11.8 The Davisson–Germer Experiment

Clinton Davisson and Lester Germer (1927) at Bell Labs in the USA gave the first direct experimental confirmation of de Broglie’s hypothesis. A heated tungsten filament $F$ emits electrons that are accelerated through a variable potential $V$ and emerge as a fine collimated beam striking a single crystal of nickel at normal incidence. A movable detector $D$ — a Faraday cylinder connected to a galvanometer — collects the scattered electrons at angle $\phi$ measured from the incident beam.

Plotting the scattered current against $\phi$ for fixed $V$, a sharp peak was observed at $\phi = 50^\circ$ when $V = 54$ V. This is exactly the angle expected for first-order Bragg diffraction from the (111) atomic planes of nickel, confirming that electrons are diffracted as waves. The de Broglie wavelength predicted by $\lambda = 12.27/\sqrt{V}$ Å at $V=54$ V is $1.67$ Å, while the wavelength inferred from the diffraction maximum is $1.65$ Å — agreement to within 1.5%. The Davisson-Germer experiment therefore established beyond doubt the wave nature of the electron. G.P. Thomson independently demonstrated electron diffraction through thin metal foils the same year; the two shared the 1937 Nobel Prize.

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11.9 Heisenberg’s Uncertainty Principle

If a particle has a wave character it cannot be localised at a mathematical point. Werner Heisenberg (1927) showed that the product of the uncertainty in position and the uncertainty in the corresponding momentum can never be smaller than a fundamental minimum:

$$\Delta x \cdot \Delta p \;\ge\; \dfrac{h}{4\pi} \;=\; \dfrac{\hbar}{2}$$

An analogous relation holds between energy and time: $\Delta E\cdot\Delta t \ge h/(4\pi)$. The principle is not a statement about the limitations of our measuring apparatus — it is a fundamental property of nature. For macroscopic objects $h$ is so small that the uncertainty is negligible, but for atomic-scale particles it is decisive.

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Textbook Exercise — Numerical Solutions

Q1. Find the maximum frequency and the minimum wavelength of X-rays produced by 30 kV electrons.

Answer: When all the kinetic energy of the electron is converted to a single photon, $h\nu_{\max}=eV$.

$$\nu_{\max}=\dfrac{eV}{h}=\dfrac{1.6\times 10^{-19}\times 30\times 10^{3}}{6.63\times 10^{-34}}=7.24\times 10^{18}\;\text{Hz}$$

$$\lambda_{\min}=\dfrac{c}{\nu_{\max}}=\dfrac{3\times 10^{8}}{7.24\times 10^{18}}=4.14\times 10^{-11}\;\text{m}=0.0414\;\text{nm}$$

Q2. The work function of caesium is 2.14 eV. (i) Find the threshold frequency. (ii) For light of frequency $6\times 10^{14}$ Hz find $K_{\max}$, $V_0$ and $v_{\max}$.

Answer:

$$\nu_0 = \dfrac{\phi_0}{h} = \dfrac{2.14\times 1.6\times 10^{-19}}{6.63\times 10^{-34}} = 5.16\times 10^{14}\;\text{Hz}$$

$$K_{\max}=h(\nu-\nu_0)=6.63\times 10^{-34}\;(6.0-5.16)\times 10^{14}=5.57\times 10^{-20}\;\text{J}=0.345\;\text{eV}$$

$$V_0=\dfrac{K_{\max}}{e}=0.345\;\text{V},\qquad v_{\max}=\sqrt{\dfrac{2K_{\max}}{m_e}}=\sqrt{\dfrac{2\times 5.57\times 10^{-20}}{9.11\times 10^{-31}}}\approx 3.49\times 10^{5}\;\text{m s}^{-1}$$

Q3. The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Answer: $K_{\max}=eV_0 = 1.6\times 10^{-19}\times 1.5 = 2.4\times 10^{-19}$ J $=1.5$ eV.

Q4. Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser of power output 9.42 mW. Find (i) the energy and momentum of each photon, (ii) the number of photons emitted per second, (iii) the speed of a hydrogen atom which has the same momentum as the photon.

$$E = \dfrac{hc}{\lambda} = \dfrac{(6.63\times 10^{-34})(3\times 10^{8})}{632.8\times 10^{-9}} = 3.14\times 10^{-19}\;\text{J} = 1.96\;\text{eV}$$

$$p = \dfrac{h}{\lambda} = \dfrac{6.63\times 10^{-34}}{632.8\times 10^{-9}} = 1.05\times 10^{-27}\;\text{kg m s}^{-1}$$

$$n = \dfrac{P}{E}=\dfrac{9.42\times 10^{-3}}{3.14\times 10^{-19}}=3.0\times 10^{16}\;\text{photons s}^{-1}$$

$$v_H = \dfrac{p}{m_H}=\dfrac{1.05\times 10^{-27}}{1.66\times 10^{-27}}=0.633\;\text{m s}^{-1}$$

Q5. The energy flux of sunlight reaching the Earth’s surface is $1.388\times 10^{3}$ W m$^{-2}$. How many photons (average $\lambda = 550$ nm) fall on 1 m$^2$ in 1 s?

$$E_\gamma = \dfrac{hc}{\lambda}=\dfrac{(6.63\times 10^{-34})(3\times 10^{8})}{550\times 10^{-9}}=3.62\times 10^{-19}\;\text{J}$$

$$n=\dfrac{1388}{3.62\times 10^{-19}}=3.84\times 10^{21}\;\text{photons m}^{-2}\;\text{s}^{-1}$$

Q6. In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency graph is found to be $4.12\times 10^{-15}$ V s. Calculate Planck’s constant.

$$h = \text{slope}\times e = (4.12\times 10^{-15})(1.6\times 10^{-19}) = 6.59\times 10^{-34}\;\text{J s}$$

Q7. A 100 W sodium lamp emits all its energy at $\lambda=589$ nm. Find (i) photon energy in eV, (ii) photons emitted per second.

$$E=\dfrac{hc}{\lambda}=3.38\times 10^{-19}\;\text{J}=2.11\;\text{eV},\quad n=\dfrac{100}{3.38\times 10^{-19}}=2.96\times 10^{20}\;\text{s}^{-1}$$

Q8. The threshold frequency for a metal is $3.3\times 10^{14}$ Hz. Light of frequency $8.2\times 10^{14}$ Hz is incident. Find the cut-off voltage.

$$V_0=\dfrac{h(\nu-\nu_0)}{e}=\dfrac{6.63\times 10^{-34}(8.2-3.3)\times 10^{14}}{1.6\times 10^{-19}}=2.03\;\text{V}$$

Q9. The work function of a metal is 4.2 eV. Will the metal show photoelectric emission for incident light of wavelength 330 nm?

$$E=\dfrac{hc}{\lambda}=\dfrac{1240}{330}\;\text{eV}=3.76\;\text{eV}<4.2\;\text{eV}$$

Since the photon energy is less than the work function, no photoelectrons are emitted.

Q10. Light of frequency $7.21\times 10^{14}$ Hz produces photoelectrons with maximum speed $6.0\times 10^{5}$ m s$^{-1}$. Find the threshold frequency for this metal.

$$K_{\max}=\tfrac12 m_e v_{\max}^2=\tfrac12(9.11\times 10^{-31})(6.0\times 10^{5})^2=1.64\times 10^{-19}\;\text{J}$$

$$\nu_0=\nu-\dfrac{K_{\max}}{h}=7.21\times 10^{14}-\dfrac{1.64\times 10^{-19}}{6.63\times 10^{-34}}=4.74\times 10^{14}\;\text{Hz}$$

Q11. Light of wavelength 488 nm is incident on the photo-cathode of a cell. The photoelectrons ejected have a stopping potential of 0.38 V. Find the work function of the cathode.

$$\phi_0=\dfrac{hc}{\lambda}-eV_0=\dfrac{1240}{488}\;\text{eV}-0.38\;\text{eV}=2.54-0.38=2.16\;\text{eV}$$

Q12. Calculate the de Broglie wavelength of (a) an electron of energy 120 eV, (b) a 0.040 kg bullet at 1000 m s$^{-1}$.

$$\text{(a)}\quad \lambda=\dfrac{12.27}{\sqrt{120}}\;\text{Å}=1.12\;\text{Å}=1.12\times 10^{-10}\;\text{m}$$

$$\text{(b)}\quad \lambda=\dfrac{h}{mv}=\dfrac{6.63\times 10^{-34}}{0.040\times 1000}=1.66\times 10^{-35}\;\text{m}$$

The macroscopic bullet has a wavelength so small that it is utterly undetectable.

Q13. The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, (b) a neutron, would have the same de Broglie wavelength.

$$K=\dfrac{p^2}{2m}=\dfrac{h^2}{2m\lambda^2}$$

$$\text{(a) electron: }K_e=\dfrac{(6.63\times 10^{-34})^2}{2(9.11\times 10^{-31})(589\times 10^{-9})^2}=6.95\times 10^{-25}\;\text{J}=4.34\times 10^{-6}\;\text{eV}$$

$$\text{(b) neutron: }K_n=\dfrac{(6.63\times 10^{-34})^2}{2(1.675\times 10^{-27})(589\times 10^{-9})^2}=3.78\times 10^{-28}\;\text{J}=2.36\times 10^{-9}\;\text{eV}$$

Q14. What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? (Atomic mass of N $=14.0076$ u.)

$$m=2(14.0076)(1.66\times 10^{-27})=4.65\times 10^{-26}\;\text{kg}$$

$$\langle K\rangle=\tfrac{3}{2}k_B T=\tfrac{3}{2}(1.38\times 10^{-23})(300)=6.21\times 10^{-21}\;\text{J}$$

$$\lambda=\dfrac{h}{\sqrt{2m\langle K\rangle}}=\dfrac{6.63\times 10^{-34}}{\sqrt{2(4.65\times 10^{-26})(6.21\times 10^{-21})}}=2.76\times 10^{-11}\;\text{m}$$

Q15. Through what potential difference must an electron be accelerated to attain a de Broglie wavelength of $1.0\times 10^{-10}$ m?

$$\lambda=\dfrac{12.27}{\sqrt{V}}\;\text{Å}\;\Rightarrow\;V=\left(\dfrac{12.27}{1.0}\right)^2=150.5\;\text{V}\approx 150\;\text{V}$$

Q16. The de Broglie wavelength of a particle of charge $1.6\times 10^{-19}$ C and mass $1.67\times 10^{-27}$ kg is $1\times 10^{-12}$ m. Find the potential difference through which it has been accelerated.

$$V=\dfrac{h^2}{2me\lambda^2}=\dfrac{(6.63\times 10^{-34})^2}{2(1.67\times 10^{-27})(1.6\times 10^{-19})(10^{-12})^2}=8.21\times 10^{5}\;\text{V}$$

Q17. Show that the de Broglie wavelength of an electron is approximately the same as that of a photon of the same energy when both have a kinetic energy of 1 keV.

$$\lambda_{e}=\dfrac{h}{\sqrt{2m_e K}}=\dfrac{6.63\times 10^{-34}}{\sqrt{2(9.11\times 10^{-31})(10^3\times 1.6\times 10^{-19})}}=3.88\times 10^{-11}\;\text{m}$$

$$\lambda_{\gamma}=\dfrac{hc}{E}=\dfrac{(6.63\times 10^{-34})(3\times 10^{8})}{10^3\times 1.6\times 10^{-19}}=1.24\times 10^{-9}\;\text{m}$$

So at 1 keV the photon wavelength is about 32 times the electron wavelength — they coincide only at much higher energies. (This question reinforces the rule $\lambda_\gamma=hc/E$ versus $\lambda_e=h/\sqrt{2mE}$.)

Q18. The position of an electron is measured with an uncertainty of $0.1$ Å. What is the minimum uncertainty in its momentum?

$$\Delta p \ge \dfrac{h}{4\pi\Delta x}=\dfrac{6.63\times 10^{-34}}{4\pi (10^{-11})}=5.27\times 10^{-24}\;\text{kg m s}^{-1}$$

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Short-Answer Questions

Q19. Define work function. State its SI unit and the unit commonly used.

Answer: The minimum energy required by an electron at the surface of a metal to escape into the vacuum is called the work function $\phi_0$. The SI unit is the joule (J), but it is conventionally expressed in electron-volt (eV); $1$ eV $=1.6\times 10^{-19}$ J.

Q20. Why does the photoelectric current saturate at a positive collector potential?

Answer: Once the collector potential is large enough that all photoelectrons emitted (regardless of their initial direction) reach the anode, the current cannot increase further; it is now limited only by the rate at which the cathode emits electrons. This rate depends on light intensity, hence the saturation current is intensity-dependent.

Q21. Why does the stopping potential remain unchanged when the intensity of light is increased keeping its frequency constant?

Answer: The stopping potential is determined by the maximum kinetic energy of the photoelectrons, which according to Einstein’s equation $K_{\max}=h\nu-\phi_0$ depends only on the frequency $\nu$ and the work function $\phi_0$. Increasing intensity raises only the number of photons (and hence the number of electrons), not the energy of any single photon.

Q22. Define threshold frequency.

Answer: The minimum frequency of incident radiation below which no photoelectric emission takes place from a given photo-sensitive surface, however large the intensity may be, is called the threshold frequency $\nu_0$. It is related to the work function by $\nu_0 = \phi_0/h$.

Q23. State two characteristic properties of a photon.

Answer: (i) A photon has zero rest mass and travels with the speed of light $c$ in vacuum. (ii) Each photon carries energy $E=h\nu$ and momentum $p=h/\lambda$ that depend only on its frequency, not on the intensity of the beam in which it occurs.

Q24. Why is the wave nature of macroscopic objects (e.g. a moving cricket ball) not observable in everyday life?

Answer: The de Broglie wavelength $\lambda = h/(mv)$ is inversely proportional to mass. For a 0.16 kg cricket ball moving at 25 m s$^{-1}$, $\lambda \approx 1.7\times 10^{-34}$ m — many orders of magnitude smaller than any conceivable obstacle (the diameter of a nucleus is $\sim 10^{-15}$ m). No diffraction or interference can be detected, so the wave aspect is unobservable.

Q25. State Heisenberg’s uncertainty principle and write its mathematical form.

Answer: The position $x$ and the momentum $p$ of a microscopic particle cannot both be measured exactly at the same instant; the product of their uncertainties is bounded below by $\Delta x \cdot \Delta p \ge h/(4\pi)$. The principle expresses an intrinsic limit imposed by the wave nature of matter, not a defect of measuring apparatus.

Q26. Distinguish between a matter wave and an electromagnetic wave.

Matter Wave	Electromagnetic Wave
Associated with a moving material particle	Associated with electric and magnetic field oscillations
Speed depends on the speed of the particle	Speed in vacuum is constant $c$
$\lambda=h/mv$	$\lambda=c/\nu$
Cannot be radiated through space	Can be radiated; carries energy and momentum through space
Wavefunction $\psi$ has no direct physical meaning; $|\psi|^2$ gives probability density	$\vec{E}$ and $\vec{B}$ are physically measurable fields

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Long-Answer / Derivation Questions

Q27. State Einstein’s photoelectric equation. Use it to explain (i) the existence of a threshold frequency, (ii) the linear dependence of $V_0$ on $\nu$, (iii) the absence of any time-lag.

Answer: Einstein assumed that radiation of frequency $\nu$ comes in indivisible quanta of energy $h\nu$. When a photon strikes an electron at the surface, the entire photon energy is transferred. Part of it ($\phi_0$) is used to overcome the surface barrier; the remainder appears as the kinetic energy of the freed electron:

$$h\nu = \phi_0 + K_{\max} = \phi_0 + \tfrac{1}{2}m v_{\max}^2$$

• Threshold frequency: Setting $K_{\max}=0$ gives $h\nu_0=\phi_0$. Below $\nu_0$, $K_{\max}$ would be negative — impossible — so no emission occurs.
• Linearity of $V_0$: Since $K_{\max}=eV_0$, the equation rearranges to $V_0 = (h/e)\nu – \phi_0/e$, a straight line of slope $h/e$ and x-intercept $\nu_0$. The slope is the same for every metal — exactly as observed.
• No time-lag: Energy transfer is a single one-photon-one-electron event taking $<10^{-9}$ s. There is no need for the electron to "accumulate" energy from successive wave crests, hence emission is essentially instantaneous.

Q28. Describe the Davisson–Germer experiment and explain how it confirmed de Broglie’s hypothesis.

Answer: Apparatus: An evacuated glass envelope contains an electron gun (heated tungsten filament $F$ and accelerating anode), a movable detector $D$ (Faraday cylinder + galvanometer), and a target consisting of a single nickel crystal. Electrons are accelerated through a variable voltage $V$ and emerge as a narrow beam striking the nickel target normal to its (111) crystal face. The detector measures the scattered electron current as a function of the scattering angle $\phi$.

Observations: (i) For low voltages no clear maximum is seen. (ii) As $V$ is increased the scattered intensity develops a sharp peak at a particular angle. (iii) For $V=54$ V the peak occurs at $\phi=50^\circ$. The atomic spacing in the (111) plane of nickel is $d=0.91$ Å, and Bragg’s condition for first-order constructive interference reads $\lambda = 2d\sin\theta$, where $\theta = (180^\circ-\phi)/2 = 65^\circ$. This gives

$$\lambda_{\text{exp}} = 2(0.91)\sin 65^\circ = 1.65\;\text{Å}$$

The de Broglie wavelength predicted for a 54 V electron is $\lambda_{\text{theory}} = 12.27/\sqrt{54}=1.67$ Å. The agreement (within 1.5%) was remarkable and beyond chance, decisively confirming that electrons exhibit wave behaviour with wavelength $h/p$ as proposed by de Broglie.

Q29. Plot a graph showing the variation of photoelectric current with collector potential for two different intensities $I_1

Both curves have the same stopping potential $V_0$ because the maximum kinetic energy depends only on $\nu$, but the saturation currents differ ($i_{s2}>i_{s1}$) because more photons (greater intensity) liberate more electrons.

Q30. Derive an expression for the de Broglie wavelength of an electron accelerated through a potential difference $V$ volts.

Answer: Kinetic energy gained $K = eV$. The momentum of the electron is

$$p = \sqrt{2 m_e K} = \sqrt{2 m_e e V}$$

$$\therefore \;\; \lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2 m_e e V}}$$

Substituting $h=6.63\times 10^{-34}$ J s, $m_e=9.11\times 10^{-31}$ kg and $e=1.6\times 10^{-19}$ C and converting to ångström,

$$\lambda \;=\; \dfrac{12.27}{\sqrt{V}}\;\text{Å}\quad(V\text{ in volt})$$

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Multiple-Choice Questions

1. The maximum kinetic energy of photoelectrons depends on:
(a) intensity (b) frequency (c) angle of incidence (d) area illuminated.
Answer: (b) frequency.

2. If the frequency of incident light is doubled, the stopping potential will:
(a) double (b) halve (c) remain the same (d) more than double.
Answer: (d) more than double, since $V_0=(h/e)\nu-\phi_0/e$ — doubling $\nu$ does more than double $V_0$ because of the negative intercept.

3. The de Broglie wavelength of a charged particle is inversely proportional to:
(a) charge (b) mass (c) momentum (d) energy.
Answer: (c) momentum.

4. The photoelectric effect can be explained on the basis of:
(a) wave theory (b) corpuscular theory (c) photon theory (d) electromagnetic theory.
Answer: (c) photon theory.

5. The Davisson-Germer experiment confirmed:
(a) particle nature of light (b) wave nature of electrons (c) existence of electron spin (d) photoelectric effect.
Answer: (b) wave nature of electrons.

6. Which of the following has the largest de Broglie wavelength when all are moving with the same kinetic energy?
(a) electron (b) proton (c) neutron (d) alpha particle.
Answer: (a) electron, since $\lambda=h/\sqrt{2mK}$ — the lightest particle has the largest wavelength.

7. The slope of the cut-off voltage vs frequency graph in the photoelectric effect is:
(a) $h$ (b) $h/e$ (c) $\phi_0$ (d) $h\nu_0$.
Answer: (b) $h/e$.

8. Which of these does not affect the work function of a metal?
(a) nature of metal (b) surface condition (c) intensity of incident light (d) impurities present.
Answer: (c) intensity of incident light.

9. Heisenberg’s uncertainty principle implies that for an electron:
(a) energy cannot be measured (b) momentum and position cannot both be known exactly (c) only velocity is uncertain (d) only mass is uncertain.
Answer: (b).

10. A photon of energy $E$ has momentum:
(a) $E/c^2$ (b) $E/c$ (c) $Ec$ (d) $E^2/c$.
Answer: (b) $E/c$.

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Fill-in-the-Blanks

1. The minimum frequency of incident light below which no photoelectrons are emitted is called the ____. (threshold frequency)
2. The energy of a photon is ____. ($h\nu$)
3. Einstein’s photoelectric equation is ____. ($h\nu=\phi_0+K_{\max}$)
4. The de Broglie wavelength of a 100 V electron is approximately ____. (1.23 Å)
5. The Davisson-Germer experiment was performed in the year ____. (1927)
6. For a photon, rest mass = ____. (zero)
7. The product of $\Delta x$ and $\Delta p$ is at least ____. ($h/4\pi$)
8. The slope of the $V_0$-$\nu$ graph is ____. ($h/e$)
9. One electron-volt = ____ joule. ($1.6\times 10^{-19}$)
10. The cut-off potential is also called the ____ potential. (stopping)

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True / False

• The photoelectric current depends on the intensity of incident light. — True.
• The kinetic energy of photoelectrons increases with intensity. — False.
• The stopping potential depends only on the work function. — False; it depends on both $\phi_0$ and $\nu$.
• A photon has zero rest mass but non-zero momentum. — True.
• Macroscopic objects show observable matter-wave behaviour. — False.
• The de Broglie wavelength of a heavier particle is smaller for the same speed. — True.
• According to classical wave theory, photoelectric emission must show a time-lag. — True (one of the failures of the classical model).
• An electron and a photon of the same energy have the same wavelength. — False.

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Additional Higher-Order Questions

Q31. Why are photoelectric experiments usually performed with alkali metals?

Answer: Alkali metals (caesium, potassium, sodium, rubidium) have low work functions of about 1.8–2.3 eV, so even visible light has enough photon energy to liberate electrons. Most other metals require ultraviolet light. Therefore alkali metals are the practical choice for photo-cathodes in light-meters, image-tubes and photovoltaic cells.

Q32. A photon and an electron have the same wavelength. Which has more momentum and which has more energy?

Answer: Both have the same momentum, since $p=h/\lambda$ holds for both. However the photon energy $E_\gamma=pc$ is far greater than the electron kinetic energy $E_e=p^2/2m_e$, because at non-relativistic speeds $E_e \ll pc$. Hence: same momentum, photon has greater energy.

Q33. What is meant by the dual nature of radiation?

Answer: Light shows wave behaviour (interference, diffraction, polarization) in some experiments and particle behaviour (photoelectric effect, Compton effect, atomic line emission) in others. The two aspects are complementary — different experimental settings reveal different aspects but never simultaneously. Light is therefore said to possess a dual nature.

Q34. State two applications of the photoelectric effect.

Answer: (i) Photoelectric cells in burglar alarms, light-operated switches, exposure meters of cameras and television-camera tubes. (ii) Photovoltaic solar cells that convert sunlight into electrical power. (iii) Optical readers in barcode scanners, smoke detectors and audio sound-tracks of older film projectors.

Q35. Compare the de Broglie wavelengths of an alpha particle and a proton accelerated through the same potential difference.

$$\lambda = \dfrac{h}{\sqrt{2mqV}}\Rightarrow\dfrac{\lambda_\alpha}{\lambda_p}=\sqrt{\dfrac{m_p q_p}{m_\alpha q_\alpha}}=\sqrt{\dfrac{m_p\cdot e}{4m_p\cdot 2e}}=\sqrt{\dfrac{1}{8}}\approx 0.354$$

Hence $\lambda_\alpha \approx \lambda_p/\sqrt{8}$ — the alpha-particle wavelength is about 35.4 % of the proton wavelength.

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Glossary of Key Terms

Term	Definition
Work function ($\phi_0$)	Minimum energy required to liberate an electron from a metal surface.
Photoelectric effect	Emission of electrons from a metal when illuminated by radiation of frequency $\nu \ge \nu_0$.
Threshold frequency ($\nu_0$)	Smallest frequency that can produce photoelectric emission from a given surface.
Threshold wavelength ($\lambda_0$)	$\lambda_0=hc/\phi_0$; longest wavelength capable of producing photoelectric emission.
Stopping potential ($V_0$)	Negative collector potential at which the photoelectric current just becomes zero.
Photon	Quantum of electromagnetic radiation; energy $h\nu$, momentum $h/\lambda$, rest mass zero.
de Broglie wavelength ($\lambda$)	Wavelength $h/p$ associated with every moving particle.
Matter wave	Probability wave whose square modulus gives the probability density of finding the particle.
Wave-particle duality	Principle that matter and radiation both possess wave and particle aspects.
Heisenberg’s principle	Lower bound $\Delta x\cdot\Delta p\ge h/(4\pi)$ on simultaneous knowledge of position and momentum.
Davisson-Germer experiment	1927 demonstration of electron diffraction from a nickel crystal — first confirmation of de Broglie’s hypothesis.
Thermionic emission	Electron emission from a hot metal surface.
Field emission	Electron emission caused by a strong external electric field.
Secondary emission	Electron emission caused by impact of energetic primary electrons or ions.
Compton effect	Increase of wavelength of a photon scattered by a free electron — corroborates the photon model.
Photo cell	Practical device based on the photoelectric effect; converts light into electrical signal.
Quantum of energy	Discrete, indivisible packet of energy $E=h\nu$ exchanged in interactions of matter and radiation.
Electron-volt (eV)	Energy gained by an electron in falling through one volt potential difference; $1$ eV $=1.6\times 10^{-19}$ J.

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Examination Tips

• Always state Einstein’s photoelectric equation in the standard form $h\nu = \phi_0 + K_{\max}$ before substituting numerical values — examiners reward the formula step.
• Convert eV to joule using $1$ eV $=1.6\times 10^{-19}$ J before equating to $hc/\lambda$ unless you keep both sides in eV (then use $hc=1240$ eV·nm).
• For the de Broglie wavelength of an electron accelerated through $V$ volts, the shortcut $\lambda=12.27/\sqrt{V}$ Å (with $V$ in volt) saves valuable time. Quote the formula explicitly before substituting.
• When asked to plot $V_0$ vs $\nu$, draw both axes, indicate the intercept $\nu_0$ on the $\nu$-axis and the negative intercept $-\phi_0/e$ on the $V_0$-axis, and label the slope $h/e$.
• For the Davisson-Germer experiment, always quote the agreement between theory ($\lambda = 1.67$ Å) and experiment ($\lambda = 1.65$ Å) at $V = 54$ V — examiners look for these numbers.
• Heisenberg’s principle: write $\Delta x\cdot\Delta p\ge h/(4\pi)$. The form $\hbar/2$ is equivalent (since $\hbar=h/2\pi$) and either is accepted.
• For numerical work, $hc = 1.24\times 10^{-6}$ eV·m $= 1240$ eV·nm — a memorisable constant that appears in almost every calculation.

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Best wishes for your ASSEB Class 12 Physics examination. Revisit this chapter with the formulas, the $V_0$-vs-$\nu$ graph and the Davisson-Germer schematic in mind — these three together cover almost every variant of the questions that appear in the board paper. Keep practising the numerical problems on photon energy, threshold frequency, stopping potential and de Broglie wavelength, and stay confident.