Class 12Physics · Modern PhysicsFull chapter

Dual Nature of Radiation and Matter

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Electron Emission and Work Function

Quick answer Free electrons in a metal are held back by the attractive pull of the positive ions; a minimum energy called the work function must be supplied to liberate them.

A metal contains a large number of free (conduction) electrons that move about randomly inside the metal but cannot normally escape its surface. This is because the surface behaves like a potential barrier — an electron trying to leave is pulled back by the net attraction of the positive ions left behind. To escape, an electron at the metal's surface must be given a minimum amount of extra energy. This minimum energy required to just eject an electron from a metal surface, without giving it any extra kinetic energy, is called the work function of that metal, denoted W0.

Work function is a property of the metal and its surface condition, and is usually expressed in electron volts (eV), where 1 eV is the energy gained by an electron accelerated through a potential difference of 1 volt (1 eV = 1.6×10-19 J). Typical work functions range from about 2 eV to 6 eV; alkali metals such as caesium, potassium and sodium have particularly low work functions (around 2-2.5 eV), which is why they are used as coatings in photosensitive devices.

Depending on how the extra energy is supplied, electron emission from a metal is classified into four types: thermionic emission, where suitable heating supplies enough thermal energy to free electrons (used in vacuum tubes and electron guns); field emission, where a very strong external electric field (~108 V/m) pulls electrons out of the metal; photoelectric emission, where light of suitable frequency shines on the surface and supplies the energy; and secondary emission, where high-speed electrons or other particles striking the surface knock out additional electrons.

Since the work function is essentially a threshold energy, it can also be expressed in terms of a minimum, or threshold frequency ν0, of electromagnetic radiation needed to eject an electron: W0 = hν0, where h is Planck's constant.

Worked Example: The work function of sodium is 2.3 eV. Find its threshold frequency and threshold wavelength.
W0 = 2.3 eV = 2.3 × 1.6×10-19 J = 3.68×10-19 J.
ν0 = W0/h = 3.68×10-19 / 6.63×10-34 = 5.55×1014 Hz.
λ0 = c/ν0 = 3×108 / 5.55×1014 = 5.41×10-7 m = 541 nm.
So light of wavelength longer than 541 nm (lower frequency) cannot eject photoelectrons from sodium no matter how intense it is.

Work Function W₀ = hν₀ J (or eV) · Minimum energy to just eject an electron from a metal surface; ν0 is the threshold frequency.
Electron Volt Conversion 1 eV = 1.6×10⁻¹⁹ J J · Convenient energy unit for atomic-scale processes.
Remember
  • Free electrons inside a metal are bound by the attractive potential of positive ions at the surface.
  • Work function W0 is the minimum energy needed to just free an electron from a metal surface.
  • Four emission types: thermionic, field, photoelectric, and secondary emission.
  • Alkali metals (Cs, K, Na) have low work functions, making them good photosensitive materials.
  • Work function and threshold frequency are related by W0 = hν0.

Photoelectric Effect: Experimental Study and Laws

Quick answer Careful experiments on light-induced electron emission revealed laws — an intensity-proportional current, a sharp threshold frequency, and instantaneous emission — that plain wave theory could not explain.

Heinrich Hertz first noticed (while studying electromagnetic waves) that sparking across a gap became easier when the electrodes were illuminated by ultraviolet light. Hallwachs and Lenard studied this effect systematically: they found that when ultraviolet light fell on a negatively charged zinc plate, the plate lost its charge, whereas a positively charged plate was unaffected — showing that negatively charged particles (electrons) were being ejected by the light.

A typical experimental set-up uses an evacuated glass/quartz tube containing a photosensitive emitter plate and a collector plate, connected to a battery and a sensitive ammeter (to measure photocurrent) with a variable, reversible potential difference applied between the plates.

Effect of intensity of light: for light of a fixed frequency (above threshold) and a fixed accelerating potential, the photoelectric current increases linearly with the intensity of incident light, eventually reaching a maximum value called the saturation current (when all emitted electrons reach the collector).

Effect of potential: if the potential of the collector is made increasingly negative with respect to the emitter, the photocurrent decreases and eventually becomes zero at a particular negative potential called the stopping potential (or cut-off potential) V0. At this potential even the most energetic photoelectrons are just turned back, so eV0 equals the maximum kinetic energy of the emitted electrons.

Effect of frequency: for a given material, the stopping potential is found to increase linearly with the frequency of incident light, but is completely independent of its intensity. Below a certain frequency, called the threshold frequency ν0 (characteristic of the material), no photoelectrons are emitted at all, however intense the light. Also, photoelectric emission is found to be an instantaneous process — electrons appear within about 10-9 s of illumination, with no measurable time lag, even for very dim light.

These observations are summarised as the laws of photoelectric effect: (i) photocurrent is directly proportional to intensity, for a fixed frequency above threshold; (ii) there exists a threshold frequency below which no emission occurs regardless of intensity; (iii) above the threshold, the maximum kinetic energy (and hence stopping potential) of photoelectrons increases linearly with frequency and is independent of intensity; (iv) emission is instantaneous, with no observable time delay.

Classical wave theory predicts that a more intense wave should simply supply more energy to each electron, so it cannot explain why Kmax is independent of intensity, why there is a threshold frequency, or why emission is instantaneous for arbitrarily weak light (energy should need time to accumulate). This failure prompted Einstein's photon explanation.

Worked Example: In an experiment, the stopping potential was found to be 0.5 V for light of frequency 5×1014 Hz and 1.5 V for frequency 7.5×1014 Hz. Find the slope of the V0 versus ν graph and hence estimate Planck's constant.
Slope = ΔV0/Δν = (1.5 - 0.5) / (7.5×1014 - 5×1014) = 1.0 / (2.5×1014) = 4.0×10-15 V·s.
Since slope = h/e, h = 4.0×10-15 × 1.6×10-19 = 6.4×10-34 J s — close to the accepted value, confirming Einstein's linear relationship.

Stopping Potential – Max KE Relation eV₀ = ½ m vₘₐₓ² = Kₘₐₓ J · The stopping potential measures the maximum kinetic energy of emitted photoelectrons.
Slope of V0 vs ν graph ΔV₀ / Δν = h / e V·s · Used experimentally to determine Planck's constant h from measured stopping potentials.
Remember
  • Hallwachs and Lenard showed UV light ejects negatively charged particles (electrons) from metal surfaces.
  • Photocurrent is directly proportional to light intensity at a fixed frequency (saturation current at high stopping voltage is zero).
  • Stopping potential V0 depends only on frequency, not intensity: eV0 = Kmax.
  • A sharp threshold frequency exists below which no photoemission occurs, regardless of intensity.
  • Photoelectric emission is instantaneous — no measurable time lag — which classical wave theory cannot explain.

Einstein's Photoelectric Equation and the Photon Picture of Light

Quick answer Einstein explained every photoelectric observation by treating light as a stream of discrete energy packets called photons, each interacting with a single electron.

Einstein proposed that light of frequency ν consists of discrete packets of energy called photons, each carrying energy E = hν, and travelling at speed c. In the photoelectric effect, a photon is absorbed entirely by a single electron in a one-to-one interaction — an electron cannot collect energy from multiple photons over time. If the photon's energy hν is less than the work function W0, the electron simply cannot escape, however many such (insufficiently energetic) photons arrive — this explains the threshold frequency. If hν ≥ W0, the electron is emitted immediately (explaining the absence of any time lag) with the leftover energy appearing as kinetic energy. Increasing the intensity of light simply increases the number of photons striking the surface per second, which increases the number of photoelectrons (current) but not the energy each one carries — this explains why Kmax is independent of intensity.

This reasoning gives Einstein's photoelectric equation: the maximum kinetic energy of an emitted photoelectron equals the photon energy minus the work function: Kmax = hν - W0 = h(ν - ν0). Since eV0 = Kmax, this can also be written as eV0 = hν - W0, which is exactly the linear relationship between stopping potential and frequency observed experimentally, with slope h/e and intercept -W0/e.

The photon picture also assigns light quanta a momentum, even though photons are massless: since energy and momentum for light are related by E = pc (as in electromagnetic theory), a photon of frequency ν (wavelength λ) has momentum p = E/c = hν/c = h/λ. This particle-like momentum is what makes light exert radiation pressure and is essential in explaining phenomena like the Compton effect.

Worked Example 1: Light of wavelength 400 nm is incident on a metal of work function 2.0 eV. Find the maximum kinetic energy of the photoelectrons and the stopping potential.
Photon energy E = hc/λ = (6.63×10-34 × 3×108) / (400×10-9) = 4.9725×10-19 J = 3.11 eV (dividing by 1.6×10-19).
Kmax = E - W0 = 3.11 - 2.0 = 1.11 eV.
Stopping potential V0 = Kmax/e = 1.11 V.

Worked Example 2: Find the momentum of a photon of wavelength 500 nm.
p = h/λ = 6.63×10-34 / (500×10-9) = 1.326×10-27 kg m/s.

Photon Energy E = hν = hc/λ J (or eV) · Energy of a single light quantum (photon) of frequency ν or wavelength λ.
Photon Momentum p = h/λ = E/c kg m/s · Momentum carried by a photon, despite it having zero rest mass.
Einstein's Photoelectric Equation Kₘₐₓ = hν − W₀ = h(ν − ν₀) J (or eV) · Maximum kinetic energy of photoelectrons in terms of incident photon energy and work function.
Stopping Potential Form eV₀ = hν − W₀ V (for V₀) · Linear relation between stopping potential and frequency; slope = h/e.
Remember
  • Light behaves as a stream of photons, each of energy E = hν, absorbed one photon per electron.
  • Einstein's equation: Kmax = hν - W0 = h(ν - ν0), matching the observed linear V0-ν graph.
  • Photon picture explains threshold frequency, intensity-independence of Kmax, and instantaneous emission all at once.
  • A photon carries momentum p = h/λ = E/c despite having zero rest mass.
  • Increasing intensity increases photon (and photoelectron) count, not photon energy.

Wave Nature of Matter: de Broglie Hypothesis

Quick answer Louis de Broglie proposed that moving matter, like light, has an associated wavelength inversely proportional to its momentum — a hypothesis that unified the wave and particle pictures.

Since light, traditionally thought of as a wave, was shown by the photoelectric effect to also have particle-like (photon) properties with momentum p = h/λ, de Broglie proposed the converse: material particles, traditionally thought of as particles, should also possess wave-like properties. He hypothesised that a particle of momentum p has an associated wavelength, called the de Broglie wavelength, given by λ = h/p = h/(mv), where m is the mass and v the speed of the particle.

Because Planck's constant h is extremely small (6.63×10-34 J s), the de Broglie wavelength of everyday macroscopic objects is immeasurably tiny and has no observable consequence, while for microscopic particles like electrons — which have very small mass — the associated wavelength can be comparable to atomic dimensions or X-ray wavelengths, making wave effects like diffraction observable.

For a charged particle of charge q and mass m accelerated from rest through a potential difference V, its kinetic energy equals qV, so p = √(2mqV), giving λ = h/√(2mqV). For an electron (q = e), this reduces to the widely used practical formula λ = 1.227/√V nm, where V is in volts.

The de Broglie hypothesis also gave a physical explanation for Bohr's quantization of angular momentum in the hydrogen atom: an electron in a stable orbit was interpreted as a standing de Broglie wave that fits exactly around the circumference, so 2πr = nλ (n = 1, 2, 3, ...). Substituting λ = h/(mv) into this condition immediately gives the Bohr quantization rule for angular momentum, L = mvr = nh/2π — showing that quantization is a natural consequence of the wave nature of the electron.

Worked Example 1: Find the de Broglie wavelength of an electron accelerated through a potential difference of 100 V.
λ = 1.227/√100 = 1.227/10 = 0.1227 nm ≈ 1.23×10-10 m.
(Check via λ = h/√(2meV): 2meV = 2 × 9.11×10-31 × 1.6×10-19 × 100 = 2.915×10-47; √(2meV) = 5.40×10-24 kg m/s; λ = 6.63×10-34/5.40×10-24 = 1.23×10-10 m = 0.123 nm — matches.)

Worked Example 2: Find the de Broglie wavelength of a 0.15 kg ball moving at 30 m/s, to see why wave effects are unobservable for macroscopic objects.
λ = h/(mv) = 6.63×10-34 / (0.15 × 30) = 6.63×10-34/4.5 = 1.47×10-34 m — vastly smaller than any measurable length, so no wave behaviour is ever observed for the ball.

de Broglie Wavelength λ = h/p = h/(mv) m · Wavelength associated with any moving particle of momentum p.
de Broglie Wavelength (via KE) λ = h/√(2mE) m · Useful when kinetic energy E of the particle is known instead of speed.
de Broglie Wavelength for Accelerated Electron λ = h/√(2meV) = 1.227/√V nm nm (V in volts) · Practical formula for an electron accelerated from rest through potential difference V.
Bohr Quantization from Standing Waves 2πr = nλ ⇒ L = mvr = nh/2π kg m²/s (for L) · de Broglie's explanation of the Bohr model's angular momentum quantization condition.
Remember
  • De Broglie hypothesis: every moving particle has an associated wavelength λ = h/p = h/(mv).
  • Wave effects are significant only for particles of very small mass (like electrons), negligible for macroscopic bodies.
  • For an electron accelerated through V volts: λ = 1.227/√V nm.
  • Bohr's angular momentum quantization L = nh/2π follows from treating electron orbits as standing de Broglie waves (2πr = nλ).
  • De Broglie waves are not electromagnetic waves; they represent the wave character of matter itself.

Davisson–Germer Experiment: Confirming Matter Waves

Quick answer The Davisson-Germer experiment showed that a beam of electrons scattered from a nickel crystal produces a diffraction pattern, providing direct experimental proof of the de Broglie hypothesis.

Clinton Davisson and Lester Germer accelerated a beam of electrons from an electron gun through a known potential difference and directed it onto the surface of a nickel crystal inside a vacuum chamber. A movable detector measured the intensity of electrons scattered at different angles from the crystal surface.

If electrons behaved purely as classical particles, the scattered intensity would vary smoothly with angle. Instead, Davisson and Germer observed a pronounced maximum in the scattered electron intensity at a specific angle for a given accelerating voltage — a diffraction pattern strikingly similar to the pattern produced when X-rays are diffracted by a crystal lattice, where the regularly spaced atomic planes act as a diffraction grating. This is analogous to the Bragg diffraction condition for waves reflecting off crystal planes, nλ = d sinθ, where d is the interplanar spacing and θ is measured from the crystal planes.

Crucially, the wavelength that had to be assigned to the electrons to explain the observed diffraction angle matched almost exactly the de Broglie wavelength λ = h/p calculated from the electron's momentum at that accelerating voltage. This quantitative agreement, obtained independently from a completely different phenomenon (diffraction, rather than any energy measurement), was the first direct experimental confirmation that matter has an intrinsic wave nature, exactly as de Broglie had hypothesised. Davisson and Germer received the Nobel Prize in Physics for this discovery.

This wave nature of fast electrons is also the physical basis of the electron microscope: since accelerated electrons can be given a de Broglie wavelength thousands of times shorter than the wavelength of visible light, electron microscopes achieve far higher resolving power than optical microscopes, allowing scientists to image structures as small as individual molecules.

Worked Example: In the Davisson-Germer experiment, electrons were accelerated through 54 V. Calculate their de Broglie wavelength and compare with the experimentally observed diffraction wavelength of about 0.165 nm.
Using λ = 1.227/√V nm: λ = 1.227/√54 = 1.227/7.348 = 0.167 nm.
This theoretical value (0.167 nm) is in excellent agreement with the wavelength (~0.165 nm) obtained independently from the electron diffraction pattern, confirming the de Broglie relation.

Bragg-type Diffraction Condition nλ = d sinθ m · Condition for constructive interference of waves (electrons or X-rays) diffracted by crystal planes of spacing d.
Remember
  • Davisson and Germer scattered electrons off a nickel crystal and observed a diffraction pattern, not smooth classical scattering.
  • The diffraction pattern is analogous to X-ray diffraction from crystal planes (Bragg-type condition nλ = d sinθ).
  • The de Broglie wavelength calculated from the accelerating voltage matched the wavelength inferred from the diffraction angle.
  • This was the first direct experimental proof of the wave nature of matter, honoured with a Nobel Prize.
  • Electron microscopes exploit the short de Broglie wavelength of fast electrons to achieve very high resolution.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

W₀ = hν₀
Work FunctionJ (or eV)
1 eV = 1.6×10⁻¹⁹ J
Electron Volt ConversionJ
eV₀ = ½ m vₘₐₓ² = Kₘₐₓ
Stopping Potential – Max KE RelationJ
ΔV₀ / Δν = h / e
Slope of V0 vs ν graphV·s
E = hν = hc/λ
Photon EnergyJ (or eV)
p = h/λ = E/c
Photon Momentumkg m/s
Kₘₐₓ = hν − W₀ = h(ν − ν₀)
Einstein's Photoelectric EquationJ (or eV)
eV₀ = hν − W₀
Stopping Potential FormV (for V₀)
λ = h/p = h/(mv)
de Broglie Wavelengthm
λ = h/√(2mE)
de Broglie Wavelength (via KE)m
λ = h/√(2meV) = 1.227/√V nm
de Broglie Wavelength for Accelerated Electronnm (V in volts)
2πr = nλ ⇒ L = mvr = nh/2π
Bohr Quantization from Standing Waveskg m²/s (for L)
nλ = d sinθ
Bragg-type Diffraction Conditionm

Test yourself

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0 correct · 0/12 answered
Q1 Photoelectric Effect - Basics easy

The phenomenon in which electrons are emitted from a metal surface when light of suitable frequency falls on it is called:

Q2 Failure of Wave Theory easy

Which observation from the photoelectric effect could NOT be explained using the classical wave theory of light?

Q3 Work Function easy

The work function of a metal is generally expressed in units of:

Q4 Stopping Potential medium

For a given photosensitive material, the stopping potential in the photoelectric effect depends on:

Q5 Work Function and Threshold Frequency medium

If W0 is the work function of a metal and h is Planck's constant, the threshold frequency ν0 is given by:

Q6 Photon Energy medium

The energy of a photon of light having wavelength 600 nm is approximately: (h = 6.63×10⁻³⁴ J s, c = 3×10⁸ m/s)

Q7 de Broglie Hypothesis medium

The de Broglie wavelength λ of a particle with momentum p is given by:

Q8 Einstein's Photoelectric Equation medium

A metal of work function 2.0 eV is illuminated by a photon of energy 3.5 eV. What is the maximum kinetic energy of the emitted photoelectron?

Q9 Stopping Potential medium

For the situation in the previous question (work function 2.0 eV, photon energy 3.5 eV), what is the stopping potential?

Q10 de Broglie Wavelength Numerical hard

What is the de Broglie wavelength of an electron accelerated from rest through a potential difference of 150 V? (Use λ = 1.227/√V nm)

Q11 Davisson-Germer Experiment hard

The Davisson-Germer experiment, in which electrons scattered from a nickel crystal produced a diffraction pattern, provided direct experimental evidence for:

Q12 Photoelectric Effect Numerical hard

Light of frequencies 5×10¹⁴ Hz and 10×10¹⁴ Hz is incident, in turn, on the same photosensitive metal. What is the difference between the two stopping potentials produced? (h = 6.63×10⁻³⁴ J s, e = 1.6×10⁻¹⁹ C)

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Monochromatic light of frequency 6.0×10¹⁴ Hz is produced by a laser. The power emitted is 2.0×10⁻³ W. Estimate the number of photons emitted per second, on average, by the source.Photon Energy and Photon Flux

Energy of each photon: E = hν = (6.63×10-34 J s)(6.0×1014 Hz) = 3.978×10-19 J.

If n photons are emitted per second, the power P = nE, so:
n = P/E = (2.0×10-3 W) / (3.978×10-19 J) = 5.03×1015 photons per second.

So the laser emits approximately 5.0×1015 photons every second on average.

2 The work function of caesium metal is 2.14 eV. When light of frequency 6×10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs. What are (a) the maximum kinetic energy of the emitted electrons, (b) the stopping potential, and (c) the maximum speed of the emitted photoelectrons?Einstein's Photoelectric Equation

Photon energy: hν = (6.63×10-34)(6×1014) = 3.978×10-19 J = 3.978×10-19/1.6×10-19 = 2.486 eV.

(a) Maximum kinetic energy:
Kmax = hν - W0 = 2.486 eV - 2.14 eV = 0.346 eV ≈ 0.35 eV (= 5.54×10-20 J).

(b) Stopping potential:
eV0 = Kmax, so V0 = 0.346 eV / e, which numerically gives V00.35 V.

(c) Maximum speed:
Kmax = ½mv²max
vmax = √(2Kmax/m) = √[2 × 5.54×10-20 / 9.11×10-31]
= √(1.216×1011) ≈ 3.49×105 m/s.

3 The photoelectric cut-off (stopping) voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?Stopping Potential and Kinetic Energy

The stopping potential relation gives the maximum kinetic energy directly:
Kmax = eV0 = e × 1.5 V = 1.5 eV.

Converting to joules: Kmax = 1.5 × 1.6×10-19 J = 2.4×10-19 J.

So the maximum kinetic energy of the emitted photoelectrons is 1.5 eV (2.4×10-19 J).

4 Monochromatic light of wavelength 632.8 nm is produced by a He-Ne laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam. (b) How many photons per second, on average, are emitted by the source?Photon Energy and Momentum

(a) Energy and momentum of each photon:
E = hc/λ = (6.63×10-34 × 3×108) / (632.8×10-9) = 1.989×10-25 / 6.328×10-7 = 3.14×10-19 J (≈ 1.96 eV).

Momentum: p = E/c = 3.14×10-19 / 3×108 = 1.05×10-27 kg m/s.

(b) Number of photons emitted per second:
n = P/E = (9.42×10-3 W) / (3.14×10-19 J) ≈ 3.0×1016 photons per second.

5 What is the de Broglie wavelength associated with an electron moving with a speed of 5.4×10⁶ m/s?de Broglie Wavelength Numerical

Using λ = h/(mv):
λ = (6.63×10-34 J s) / [(9.11×10-31 kg)(5.4×106 m/s)]
= 6.63×10-34 / 4.919×10-24

λ ≈ 1.35×10-10 m = 0.135 nm.

This wavelength is comparable to atomic/crystal-plane spacings, which is why electron diffraction is observable, unlike for macroscopic objects.

6 An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of the electron.Comparing Photon and Matter-wave Properties

(a) Momentum (same for both, since p = h/λ):
p = h/λ = (6.63×10-34) / (1.00×10-9) = 6.63×10-25 kg m/s.

(b) Energy of the photon:
Ephoton = pc = (6.63×10-25)(3×108) = 1.989×10-16 J = 1.989×10-16/1.6×10-191243 eV (≈1.24 keV).

(c) Kinetic energy of the electron:
Ke = p²/2m = (6.63×10-25)² / (2 × 9.11×10-31) = 4.396×10-49 / 1.822×10-30 = 2.413×10-19 J

= 2.413×10-19/1.6×10-191.51 eV.

Note that for the same wavelength (same momentum), the photon carries far more energy (~1.24 keV) than the electron's kinetic energy (~1.51 eV), because the photon's energy is pc while the electron's kinetic energy is p²/2m — a consequence of the electron having rest mass and moving non-relativistically.

Previous-year board questions 4

Q1 Derive Einstein's photoelectric equation. Using it, explain (i) why photoelectric current is independent of the frequency of incident radiation above threshold, and (ii) how the stopping potential varies linearly with frequency of incident light. 2023 3 marks

Derivation: According to Einstein, light of frequency ν consists of photons each of energy E = hν. In the photoelectric process, a single photon is absorbed by a single electron. Part of this energy, equal to the work function W0, is used to free the electron from the metal surface, and the rest appears as the kinetic energy of the emitted electron. For the fastest (least tightly bound, surface) electrons, this gives the maximum kinetic energy:

Kmax = hν - W0

Since the maximum kinetic energy is related to the stopping potential V0 by Kmax = eV0, we get:

eV0 = hν - W0 ⇒ V0 = (h/e)ν - W0/e

This is Einstein's photoelectric equation.

(i) The photocurrent (i.e., the number of photoelectrons emitted per second) depends on the number of photons incident per second, which is determined by the intensity of light, not its frequency. Increasing frequency (at constant intensity) increases the energy per photon but not necessarily the number of photons, so photocurrent is essentially governed by intensity, not frequency, provided ν > ν0.

(ii) From V0 = (h/e)ν - W0/e, V0 is a linear function of ν with slope h/e (a universal constant, same for all metals) and intercept -W0/e (which depends on the metal). This matches the experimentally observed straight-line graphs of stopping potential versus frequency for different metals, all having the same slope but different intercepts.

Q2 The threshold frequency for photoelectric emission from a certain metal surface is 3.3×10¹⁴ Hz. If light of frequency 8.2×10¹⁴ Hz is incident on this surface, calculate the cut-off (stopping) voltage for the photoelectrons. 2022 2 marks

Given: ν0 = 3.3×1014 Hz, ν = 8.2×1014 Hz.

By Einstein's photoelectric equation:
eV0 = h(ν - ν0) = (6.63×10-34)(8.2×1014 - 3.3×1014)
= (6.63×10-34)(4.9×1014)
= 3.249×10-19 J

V0 = 3.249×10-19 / 1.6×10-192.03 V ≈ 2.0 V.

So the stopping voltage required is approximately 2.0 V.

Q3 An electron and a proton are accelerated through the same potential difference from rest. Find the ratio of their de Broglie wavelengths (λe : λp). (mp ≈ 1836 me) 2023 3 marks

For a charge q of mass m accelerated from rest through potential difference V, kinetic energy KE = qV = p²/2m, so momentum p = √(2mqV).

De Broglie wavelength: λ = h/p = h/√(2mqV).

Since both electron and proton carry the same magnitude of charge e and are accelerated through the same V:

λep = √(2mpeV) / √(2meeV) = √(mp/me)

Given mp/me ≈ 1836:
λep = √1836 ≈ 42.8

So λe : λp ≈ 42.8 : 1 — the electron (lighter particle) has a much longer de Broglie wavelength than the proton for the same accelerating potential, since lower mass gives lower momentum for the same kinetic energy.

Q4 (a) Define stopping potential and threshold frequency in the context of the photoelectric effect. (b) Sketch and describe the nature of the graph between stopping potential (V0) and frequency (ν) of incident radiation for two different photosensitive metals having work functions W1 and W2, with W1 > W2. 2020 3 marks

Stopping potential (V0): It is the minimum negative (retarding) potential applied to the collector plate, relative to the emitter, at which the photoelectric current becomes zero — it just stops even the most energetic photoelectrons from reaching the collector. It measures the maximum kinetic energy of the photoelectrons: eV0 = Kmax.

Threshold frequency (ν0): It is the minimum frequency of incident light below which no photoelectric emission occurs from a given metal surface, no matter how intense the light is; ν0 = W0/h.

Graph description: For each metal, a plot of V0 (y-axis) against ν (x-axis) is a straight line described by V0 = (h/e)ν - W0/e. Both lines have the same slope h/e (a universal constant independent of the metal), so the two lines are parallel. However, since metal 1 has a larger work function (W1 > W2), its line is shifted to the right/down: it has a larger x-intercept (larger threshold frequency ν01 = W1/h) and a more negative y-intercept (-W1/e), compared to metal 2's line, which starts emitting at a lower threshold frequency ν02 = W2/h and has a less negative y-intercept. Both lines converge to the same slope, illustrating that h/e is a universal constant while W0 (and hence ν0) is metal-specific.

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