Dual nature of radiation
Light travels and interferes like a wave, but exchanges energy and momentum in localized photons. The experiment decides which model is useful.
Dual Nature Formulas PYQs
Dual Nature of Radiation and Matter Formulae & PYQs
Dual Nature Formulas PYQs complete revision page with formula sheet, NCERT solutions, NEET questions, JEE Main questions, JEE Advanced difficult questions, IB, ICSE, IGCSE and A-Level practice.
37NCERT solutions
120+fresh practice questions
8clean SVG diagrams
10exam mistake alerts
01
Concept overview
Quick Concept Overview
Dual Nature Formulas PYQs are easiest when the chapter is first understood as a clean map of photons, photoelectric graphs, matter waves and diffraction.
Photon theory
A photon of frequency ν has energy hν and momentum h/λ. Increasing intensity increases photon rate; increasing frequency increases energy per photon.
Photoelectric effect
Electrons are emitted from a metal only when photon energy exceeds the work function. Emission is almost instantaneous and Kmax depends on frequency.
Einstein photoelectric equation
The equation hν = Φ + Kmax explains threshold frequency, stopping potential, and the slope of Kmax versus ν graphs.
de Broglie wavelength
A moving particle of momentum p has wavelength h/p. Atomic-scale wavelengths make electron diffraction and electron microscopes possible.
Davisson-Germer experiment
Electron diffraction from a nickel crystal showed a sharp intensity maximum. The Bragg wavelength matched the de Broglie wavelength.
Matter waves
Matter-wave behavior is visible when λ is comparable with slit width, aperture size, or interplanar spacing. For macroscopic bodies λ is too small to observe.
02
Formula sheet
Dual Nature Formula Sheet
High-frequency formulas with exam-use notes and unit cautions.
Photon energy
E = hν
Energy increases with frequency. Frequency, not intensity, decides whether one photon can eject one electron.
Wavelength form
E = hc/λ
Use E(eV) = 1240/λ(nm) for fast numerical work.
Einstein equation
Kmax = hν - Φ
Maximum kinetic energy is the excess photon energy after overcoming the work function.
Stopping potential
eV0 = Kmax
A retarding potential just stopping the fastest photoelectrons measures Kmax.
Threshold frequency
ν0 = Φ/h
Below this frequency no photoelectric emission occurs, however large the intensity is.
Threshold wavelength
λ0 = hc/Φ
This is the longest wavelength that can produce photoemission.
de Broglie relation
λ = h/p
Every moving particle has an associated matter wavelength.
Non-relativistic particle
λ = h/mv
Use when speed is well below c.
Kinetic-energy form
λ = h/√(2mK)
Useful when kinetic energy is given directly.
Electron through voltage
λ = h/√(2meeV)
For electrons accelerated from rest through potential difference V.
Electron shortcut
λ(Å) = 12.27/√V
V must be in volts and λ comes out in angstroms.
Bragg law
nλ = 2d sinθ
Use the glancing angle θ with the crystal plane, not blindly the scattering angle.
03
Diagrams
Clean SVG Diagrams for Dual Nature
Photoelectric effect, graph interpretation, matter waves, Davisson-Germer and Bragg diffraction.
04
NCERT solutions
NCERT Questions with Complete Solutions
All NCERT Exercise 11.1 to 11.37 retained as visible HTML with clearer steps, formulas and final answers.
NCERT 11.1Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.
Given Data
Accelerating voltage V = 30 kV = 3.0×10⁴ V; electron charge e = 1.602×10⁻¹⁹ C.
Formula Used
Maximum photon energy: hνmax = eV; minimum wavelength: λmin = hc/(eV).
Step-by-Step Solution
The maximum energy of an X-ray photon equals the complete kinetic energy of one electron.
Emax = eV = 30,000 eV = 4.806×10⁻¹⁵ J.
νmax = E/h = (4.806×10⁻¹⁵)/(6.626×10⁻³⁴) = 7.25×10¹⁸ Hz.
λmin = c/νmax = (3.00×10⁸)/(7.25×10¹⁸) = 4.14×10⁻¹¹ m.
Final Answer
(a) νmax = 7.25×10¹⁸ Hz; (b) λmin = 4.14×10⁻¹¹ m = 0.414 Å.
Exam Tip
For an X-ray tube, use the full electron energy eV for the short-wavelength limit.
NCERT 11.2The 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 is the (a) maximum kinetic energy of the emitted electrons, (b) stopping potential, and (c) maximum speed of the emitted photoelectrons?
Given Data
Φ = 2.14 eV; ν = 6×10¹⁴ Hz; h = 6.626×10⁻³⁴ J s; me = 9.11×10⁻³¹ kg.
Formula Used
Kmax = hν−Φ; eV₀ = Kmax; Kmax = ½mev².
Step-by-Step Solution
Photon energy hν = 6.626×10⁻³⁴×6×10¹⁴ = 3.976×10⁻¹⁹ J = 2.48 eV.
Kmax = 2.48−2.14 = 0.34 eV = 5.45×10⁻²⁰ J.
V₀ = Kmax/e = 0.34 V.
v = √(2K/m) = √[2(5.45×10⁻²⁰)/(9.11×10⁻³¹)] = 3.46×10⁵ m s⁻¹.
Final Answer
(a) 0.34 eV = 5.45×10⁻²⁰ J; (b) 0.34 V; (c) 3.46×10⁵ m s⁻¹.
Exam Tip
When kinetic energy is expressed in eV, its numerical value directly gives the stopping potential in volts.
NCERT 11.3The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Given Data
Stopping potential V₀ = 1.5 V.
Formula Used
Kmax = eV₀.
Step-by-Step Solution
Kmax = (1.602×10⁻¹⁹ C)(1.5 V) = 2.403×10⁻¹⁹ J. In electron-volts, Kmax = 1.5 eV.
Final Answer
Kmax = 1.5 eV = 2.40×10⁻¹⁹ J.
Exam Tip
Do not confuse stopping potential in volts with energy in joules; multiply by e for SI energy.
NCERT 11.4Monochromatic light of wavelength 632.8 nm is produced by a helium-neon 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 the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area.) (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Given Data
λ = 632.8 nm = 6.328×10⁻⁷ m; P = 9.42×10⁻³ W; mH = 1.67×10⁻²⁷ kg.
Formula Used
E = hc/λ; p = h/λ; N = P/E; v = p/mH.
Step-by-Step Solution
E = (6.626×10⁻³⁴×3×10⁸)/(6.328×10⁻⁷) = 3.14×10⁻¹⁹ J = 1.96 eV.
p = 6.626×10⁻³⁴/(6.328×10⁻⁷) = 1.047×10⁻²⁷ kg m s⁻¹.
N = 9.42×10⁻³/(3.14×10⁻¹⁹) = 3.00×10¹⁶ photons s⁻¹.
vH = (1.047×10⁻²⁷)/(1.67×10⁻²⁷) = 0.627 m s⁻¹.
Final Answer
(a) E = 3.14×10⁻¹⁹ J and p = 1.047×10⁻²⁷ kg m s⁻¹; (b) 3.00×10¹⁶ photons s⁻¹; (c) 0.627 m s⁻¹.
Exam Tip
Power divided by energy per photon gives the photon arrival rate.
NCERT 11.5The energy flux of sunlight reaching the surface of the earth is (1.388 × 10^3) W m-2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Given Data
Intensity I = 1.388×10³ W m⁻²; λ = 550 nm.
Formula Used
Ephoton = hc/λ; photon flux = I/Ephoton.
Step-by-Step Solution
E = (6.626×10⁻³⁴×3×10⁸)/(550×10⁻⁹) = 3.61×10⁻¹⁹ J.
Photon flux = 1.388×10³/(3.61×10⁻¹⁹) = 3.84×10²¹ photons m⁻² s⁻¹.
Final Answer
Approximately 3.8×10²¹ photons m⁻² s⁻¹.
Exam Tip
Energy flux is power per unit area, so the result is a photon rate per unit area.
NCERT 11.6In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be (4.12 × 10-15) V s. Calculate the value of Planck's constant.
Given Data
Slope of V₀ versus ν = 4.12×10⁻¹⁵ V s.
Formula Used
From eV₀ = hν−Φ, slope = h/e; therefore h = e×slope.
Step-by-Step Solution
h = (1.602×10⁻¹⁹ C)(4.12×10⁻¹⁵ V s) = 6.60×10⁻³⁴ J s.
Final Answer
h = 6.60×10⁻³⁴ J s.
Exam Tip
For a V₀–ν graph, slope is h/e; for a Kmax–ν graph, slope is h.
NCERT 11.7A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?
Given Data
P = 100 W; λ = 589 nm.
Formula Used
E = hc/λ; N = P/E.
Step-by-Step Solution
E = (6.626×10⁻³⁴×3×10⁸)/(589×10⁻⁹) = 3.375×10⁻¹⁹ J.
N = 100/(3.375×10⁻¹⁹) = 2.96×10²⁰ photons s⁻¹.
Final Answer
(a) 3.38×10⁻¹⁹ J per photon; (b) 2.96×10²⁰ photons s⁻¹.
Exam Tip
Because the whole sphere absorbs the radiation, no inverse-square area factor is needed for the total rate.
NCERT 11.8The threshold frequency for a certain metal is (3.3 × 1014) Hz. If light of frequency (8.2 × 1014) Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Given Data
ν₀ = 3.3×10¹⁴ Hz; ν = 8.2×10¹⁴ Hz.
Formula Used
eV₀ = h(ν−ν₀), or V₀ = (h/e)(ν−ν₀).
Step-by-Step Solution
ν−ν₀ = 4.9×10¹⁴ Hz.
V₀ = 4.136×10⁻¹⁵×4.9×10¹⁴ = 2.03 V.
Final Answer
Cut-off voltage V₀ ≈ 2.0 V.
Exam Tip
Only the excess frequency above threshold contributes to photoelectron kinetic energy.
NCERT 11.9The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Given Data
Φ = 4.2 eV; λ = 330 nm.
Formula Used
Photon energy E(eV) = 1240/λ(nm). Emission requires E ≥ Φ.
Step-by-Step Solution
E = 1240/330 = 3.76 eV. Since 3.76 eV < 4.2 eV, a photon cannot liberate an electron.
Final Answer
No photoelectric emission occurs.
Exam Tip
Increasing intensity cannot compensate for photon energy below the work function.
NCERT 11.10Light of frequency (7.21 × 1014) Hz is incident on a metal surface. Electrons with a maximum speed of (6.0 × 10^5) m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Given Data
ν = 7.21×10¹⁴ Hz; vmax = 6.0×10⁵ m s⁻¹.
Formula Used
hν = hν₀ + ½mv²; ν₀ = ν−mv²/(2h).
Step-by-Step Solution
Kmax = ½(9.11×10⁻³¹)(6×10⁵)² = 1.64×10⁻¹⁹ J.
K/h = (1.64×10⁻¹⁹)/(6.626×10⁻³⁴) = 2.47×10¹⁴ Hz.
ν₀ = 7.21×10¹⁴−2.47×10¹⁴ = 4.74×10¹⁴ Hz.
Final Answer
Threshold frequency ν₀ = 4.74×10¹⁴ Hz.
Exam Tip
Use the maximum photoelectron speed because Einstein's equation uses Kmax.
NCERT 11.11Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Given Data
λ = 488 nm; V₀ = 0.38 V.
Formula Used
Φ = hc/λ−eV₀, or Φ(eV) = 1240/λ(nm)−V₀.
Step-by-Step Solution
Photon energy = 1240/488 = 2.541 eV.
Maximum kinetic energy = eV₀ = 0.38 eV.
Φ = 2.541−0.38 = 2.161 eV.
Final Answer
Work function Φ ≈ 2.16 eV = 3.46×10⁻¹⁹ J.
Exam Tip
In eV units, subtract the stopping potential numerically from photon energy in eV.
NCERT 11.12Calculate (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Given Data
V = 56 V; me = 9.11×10⁻³¹ kg.
Formula Used
p = √(2meeV); λ = h/p = 12.27/√V Å.
Step-by-Step Solution
p = √[2(9.11×10⁻³¹)(1.602×10⁻¹⁹)(56)] = 4.04×10⁻²⁴ kg m s⁻¹.
λ = 6.626×10⁻³⁴/(4.04×10⁻²⁴) = 1.64×10⁻¹⁰ m = 1.64 Å.
Final Answer
(a) p = 4.04×10⁻²⁴ kg m s⁻¹; (b) λ = 1.64×10⁻¹⁰ m.
Exam Tip
The shortcut 12.27/√V gives electron wavelength directly in angstroms.
NCERT 11.13What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV?
Given Data
K = 120 eV = 1.922×10⁻¹⁷ J.
Formula Used
p = √(2mK); v = √(2K/m); λ = h/p.
Step-by-Step Solution
p = √[2(9.11×10⁻³¹)(1.922×10⁻¹⁷)] = 5.92×10⁻²⁴ kg m s⁻¹.
v = p/m = 6.50×10⁶ m s⁻¹.
λ = 6.626×10⁻³⁴/(5.92×10⁻²⁴) = 1.12×10⁻¹⁰ m.
Final Answer
(a) 5.92×10⁻²⁴ kg m s⁻¹; (b) 6.50×10⁶ m s⁻¹; (c) 1.12×10⁻¹⁰ m.
Exam Tip
At 120 eV the electron is safely non-relativistic.
NCERT 11.14The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de Broglie wavelength.
Given Data
λ = 589 nm = 5.89×10⁻⁷ m; me = 9.11×10⁻³¹ kg; mn = 1.675×10⁻²⁷ kg.
Formula Used
K = p²/(2m) = h²/(2mλ²).
Step-by-Step Solution
Common momentum p = h/λ = 1.125×10⁻²⁷ kg m s⁻¹.
Electron: Ke = p²/(2me) = 6.95×10⁻²⁵ J = 4.34×10⁻⁶ eV.
Neutron: Kn = p²/(2mn) = 3.78×10⁻²⁸ J = 2.36×10⁻⁹ eV.
Final Answer
(a) Electron: 6.95×10⁻²⁵ J; (b) neutron: 3.78×10⁻²⁸ J.
Exam Tip
At equal wavelength particles have equal momentum, but the heavier particle has less kinetic energy.
NCERT 11.15What is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at a speed of 1.0 km/s, (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and (c) a dust particle of mass (1.0 × 10-9) kg drifting with a speed of 2.2 m/s?
Given Data
Use the three stated masses and speeds.
Formula Used
λ = h/(mv).
Step-by-Step Solution
(a) p = 0.040×1000 = 40 kg m s⁻¹; λ = 6.626×10⁻³⁴/40 = 1.66×10⁻³⁵ m.
(b) p = 0.060×1.0 = 0.060; λ = 1.10×10⁻³² m.
(c) p = 1.0×10⁻⁹×2.2 = 2.2×10⁻⁹; λ = 3.01×10⁻²⁵ m.
Final Answer
(a) 1.66×10⁻³⁵ m; (b) 1.10×10⁻³² m; (c) 3.01×10⁻²⁵ m.
Exam Tip
Macroscopic wavelengths are far too small for observable diffraction.
NCERT 11.16An 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.
Given Data
λ = 1.00 nm = 1.00×10⁻⁹ m.
Formula Used
p = h/λ; Eph = hc/λ; Ke = p²/(2me).
Step-by-Step Solution
Both have p = 6.626×10⁻³⁴/10⁻⁹ = 6.626×10⁻²⁵ kg m s⁻¹.
Eph = pc = 1.988×10⁻¹⁶ J = 1.24 keV.
Ke = (6.626×10⁻²⁵)²/[2(9.11×10⁻³¹)] = 2.41×10⁻¹⁹ J = 1.50 eV.
Final Answer
(a) 6.626×10⁻²⁵ kg m s⁻¹ each; (b) 1.988×10⁻¹⁶ J; (c) 2.41×10⁻¹⁹ J.
Exam Tip
Equal wavelength means equal momentum, not equal energy.
NCERT 11.17(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be (1.40 × 10-10) m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of ((3/2)kT) at 300 K.
Given Data
mn = 1.675×10⁻²⁷ kg; λ = 1.40×10⁻¹⁰ m; T = 300 K.
Formula Used
K = h²/(2mλ²); for thermal neutron K = 3kT/2 and λ = h/√(2mK) = h/√(3mkT).
Step-by-Step Solution
(a) K = (6.626×10⁻³⁴)²/[2(1.675×10⁻²⁷)(1.40×10⁻¹⁰)²] = 6.69×10⁻²¹ J = 0.0418 eV.
(b) K = 1.5(1.381×10⁻²³)(300) = 6.21×10⁻²¹ J.
λ = 6.626×10⁻³⁴/√[2(1.675×10⁻²⁷)(6.21×10⁻²¹)] = 1.45×10⁻¹⁰ m.
Final Answer
(a) 6.69×10⁻²¹ J = 0.0418 eV; (b) 1.45×10⁻¹⁰ m.
Exam Tip
Thermal neutron wavelengths are comparable with interatomic spacing, making them useful diffraction probes.
NCERT 11.18Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Given Data
Photon energy E = hν and relativistic photon relation E = pc.
Formula Used
λdB = h/p; electromagnetic wavelength λ = c/ν.
Step-by-Step Solution
For a photon, hν = pc, so p = hν/c. Since ν/c = 1/λ, p = h/λ. Therefore λdB = h/p = h/(h/λ) = λ.
Final Answer
The photon's de Broglie wavelength equals the wavelength of its electromagnetic radiation.
Exam Tip
For photons never use K = p²/(2m); use E = pc.
NCERT 11.19What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Given Data
Molecular mass N₂ = 28.0152 u = 4.652×10⁻²⁶ kg; T = 300 K.
Formula Used
vrms = √(3kT/m); λ = h/(mvrms) = h/√(3mkT).
Step-by-Step Solution
vrms = √[3(1.381×10⁻²³)(300)/(4.652×10⁻²⁶)] = 517 m s⁻¹.
λ = 6.626×10⁻³⁴/[4.652×10⁻²⁶×517] = 2.76×10⁻¹¹ m.
Final Answer
λ ≈ 2.76×10⁻¹¹ m.
Exam Tip
Use molecular mass 28 u, not the atomic mass 14 u.
NCERT 11.20(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its (e/m), is given to be (1.76 × 1011) C kg-1. (b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Given Data
e/m = 1.76×10¹¹ C kg⁻¹; V₁ = 500 V; V₂ = 10⁷ V.
Formula Used
Non-relativistic: v = √[2(e/m)V]. Relativistic: eV = (γ−1)mc² and v = c√[1−1/γ²].
Step-by-Step Solution
(a) v = √[2(1.76×10¹¹)(500)] = 1.33×10⁷ m s⁻¹.
(b) Classical formula gives v = √[2(1.76×10¹¹)(10⁷)] = 1.88×10⁹ m s⁻¹, which exceeds c and is impossible.
For 10 MeV, γ = 1 + 10/0.511 = 20.57. Hence v = c√(1−1/20.57²) ≈ 0.9988c.
Final Answer
(a) 1.33×10⁷ m s⁻¹; (b) the classical answer is invalid; relativistically v ≈ 0.9988c.
Exam Tip
Any computed speed comparable with or greater than c signals the need for relativistic mechanics.
NCERT 11.21(a) A monoenergetic electron beam with electron speed of (5.20 × 10^6) m s-1 is subject to a magnetic field of (1.30 × 10-4) T normal to the beam velocity. What is the radius of the circle traced by the beam, given (e/m) for electron equals (1.76 × 1011) C kg-1? (b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron? If not, in what way is it modified?
Given Data
v = 5.20×10⁶ m s⁻¹; B = 1.30×10⁻⁴ T.
Formula Used
Non-relativistic r = mv/(eB) = v/[(e/m)B]. Relativistic r = p/(eB) = γmv/(eB).
Step-by-Step Solution
(a) r = 5.20×10⁶/[(1.76×10¹¹)(1.30×10⁻⁴)] = 0.227 m.
(b) At 20 MeV, K is much larger than the 0.511 MeV rest energy, so p≠mv. Replace mv by relativistic momentum γmv, or obtain p from E² = p²c² + m²c⁴.
Final Answer
(a) r = 0.227 m; (b) use r = p/(eB) with relativistic momentum.
Exam Tip
The magnetic force law remains valid; it is the momentum-speed relation that changes.
NCERT 11.22An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure ((10-2) mm of Hg). A magnetic field of (2.83 × 10-4) T curves the path of the electrons in a circular orbit of radius 12.0 cm. Determine (e/m) from the data.
Given Data
V = 100 V; B = 2.83×10⁻⁴ T; r = 0.120 m.
Formula Used
eV = ½mv² and evB = mv²/r, giving e/m = 2V/(B²r²).
Step-by-Step Solution
e/m = 2(100)/[(2.83×10⁻⁴)²(0.120)²].
B²r² = 1.153×10⁻⁹.
e/m = 200/(1.153×10⁻⁹) = 1.73×10¹¹ C kg⁻¹.
Final Answer
e/m ≈ 1.73×10¹¹ C kg⁻¹.
Exam Tip
Convert radius to metres before substitution.
NCERT 11.23(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation? (b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Given Data
λmin = 0.45 Å = 4.5×10⁻¹¹ m.
Formula Used
Emax = hc/λmin; eV = Emax.
Step-by-Step Solution
E = (6.626×10⁻³⁴×3×10⁸)/(4.5×10⁻¹¹) = 4.42×10⁻¹⁵ J.
In eV: E = 4.42×10⁻¹⁵/(1.602×10⁻¹⁹) = 2.76×10⁴ eV = 27.6 keV.
Therefore V ≈ 27.6 kV.
Final Answer
(a) 4.42×10⁻¹⁵ J = 27.6 keV; (b) about 28 kV.
Exam Tip
Numerically, an electron accelerated through 1 kV gains 1 keV.
NCERT 11.24In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1 BeV = (10^9) eV)
Given Data
Total energy = 10.2×10⁹ eV; each photon gets 5.1×10⁹ eV.
Formula Used
λ = hc/E; hc = 1240 eV nm.
Step-by-Step Solution
λ = 1240/(5.1×10⁹) nm = 2.43×10⁻⁷ nm = 2.43×10⁻¹⁶ m.
Final Answer
λ ≈ 2.43×10⁻¹⁶ m for each γ-ray.
Exam Tip
Equal-energy two-photon annihilation conserves momentum because the photons travel in opposite directions.
NCERT 11.25Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light. (a) The number of photons emitted per second by a Medium Wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m. (b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that humans can perceive ((~ 10-10) W m-2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about (6 × 1014) Hz.
Given Data
(a) P = 10⁴ W, λ = 500 m. (b) I = 10⁻¹⁰ W m⁻², A = 0.4 cm² = 4×10⁻⁵ m², ν = 6×10¹⁴ Hz.
Formula Used
N = P/(hc/λ); for the eye P = IA and N = P/(hν).
Step-by-Step Solution
(a) Photon energy = hc/λ = 3.976×10⁻²⁸ J. N = 10⁴/(3.976×10⁻²⁸) = 2.52×10³¹ s⁻¹.
(b) P entering pupil = 10⁻¹⁰×4×10⁻⁵ = 4×10⁻¹⁵ W. Photon energy = 6.626×10⁻³⁴×6×10¹⁴ = 3.976×10⁻¹⁹ J. N = 4×10⁻¹⁵/3.976×10⁻¹⁹ ≈ 1.0×10⁴ s⁻¹.
Final Answer
(a) 2.52×10³¹ photons s⁻¹; (b) approximately 10⁴ photons s⁻¹.
Exam Tip
Low-frequency radio photons have tiny individual energy, so ordinary radio power contains enormous photon numbers.
NCERT 11.26Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is −1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity ((~ 10^5) W m-2) red light of wavelength 6328 Å produced by a He-Ne laser?
Given Data
λUV = 227.1 nm; |V₀| = 1.3 V; λred = 632.8 nm.
Formula Used
Φ = hc/λ−e|V₀|; emission requires hc/λ ≥ Φ.
Step-by-Step Solution
UV photon energy = 1240/227.1 = 5.46 eV.
Φ = 5.46−1.30 = 4.16 eV.
Red photon energy = 1240/632.8 = 1.96 eV, which is below 4.16 eV. Therefore no photoemission occurs, even at very high intensity.
Final Answer
Work function ≈ 4.16 eV; the red laser produces no photoelectric emission.
Exam Tip
Intensity changes photon number, not energy per photon.
NCERT 11.27Monochromatic radiation of wavelength 640.2 nm from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Given Data
λ₁ = 640.2 nm; V₀₁ = 0.54 V; λ₂ = 427.2 nm.
Formula Used
Φ = hc/λ₁−eV₀₁; V₀₂ = hc/(eλ₂)−Φ/e.
Step-by-Step Solution
Φ = 1240/640.2−0.54 = 1.397 eV.
New photon energy = 1240/427.2 = 2.903 eV.
V₀₂ = 2.903−1.397 = 1.506 V.
Final Answer
New stopping voltage ≈ 1.51 V.
Exam Tip
The work function stays unchanged because the photocathode material is unchanged.
NCERT 11.28A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In an experiment with rubidium photo-cell, the following lines from a mercury source were used: λ_1 = 3650 Å,; λ_2 = 4047 Å,; λ_3 = 4358 Å,; λ_4 = 5461 Å,; λ_5 = 6907 Å The stopping voltages, respectively, were measured to be: V_{01}=1.28V,; V_{02}=0.95V,; V_{03}=0.74V,; V_{04}=0.16V,; V_{05}=0V Determine the value of Planck's constant (h), the threshold frequency and work function for the material.
Given Data
Five pairs of wavelength and stopping voltage are supplied.
Formula Used
V₀ = (h/e)ν−Φ/e, with ν = c/λ. Slope of V₀–ν graph is h/e; x-intercept is ν₀; Φ = hν₀.
Step-by-Step Solution
Convert each wavelength to frequency and plot V₀ against ν. Representative values are ν = 8.22, 7.41, 6.88, 5.49 and 4.34 ×10¹⁴ Hz.
Using the best-fit straight line, slope ≈ 4.12×10⁻¹⁵ V s.
h = e×slope = 1.602×10⁻¹⁹×4.12×10⁻¹⁵ ≈ 6.60×10⁻³⁴ J s.
The line cuts V₀ = 0 near ν₀ ≈ 5.1×10¹⁴ Hz.
Φ = hν₀ ≈ 3.37×10⁻¹⁹ J ≈ 2.1 eV.
Final Answer
h ≈ 6.6×10⁻³⁴ J s; ν₀ ≈ 5.1×10¹⁴ Hz; Φ ≈ 2.1 eV.
Exam Tip
Use a best-fit line rather than relying on the 6907 Å zero reading alone.
NCERT 11.29The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Given Data
λ = 3300 Å = 330 nm; listed work functions.
Formula Used
Photon energy E = 1240/λ(nm); emission occurs if E ≥ Φ.
Step-by-Step Solution
E = 1240/330 = 3.76 eV.
Na and K have Φ below 3.76 eV, so they emit. Mo and Ni have Φ above 3.76 eV, so they do not emit.
At 50 cm the intensity becomes four times larger, but photon energy remains 3.76 eV. Hence the emitting/non-emitting metals are unchanged; only photocurrent from Na and K increases.
Final Answer
Mo and Ni do not emit at either distance; Na and K emit, with greater current at 50 cm.
Exam Tip
Distance affects intensity and photocurrent, not threshold frequency or stopping potential.
NCERT 11.30Light of intensity (10-5) W m-2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate the time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Given Data
I = 10⁻⁵ W m⁻²; A = 2×10⁻⁴ m²; Φ = 2 eV ≈ 3.2×10⁻¹⁹ J; approximately 10¹⁷ atoms in five illuminated layers.
Formula Used
Total power P = IA; power per atom = P/N; accumulation time t = Φ/(P/N).
Step-by-Step Solution
P = 10⁻⁵×2×10⁻⁴ = 2×10⁻⁹ W.
Taking an atomic area of order 10⁻²⁰ m² gives roughly 2×10¹⁶ atoms per layer, or N≈10¹⁷ atoms in five layers.
Power per atom ≈ 2×10⁻⁹/10¹⁷ = 2×10⁻²⁶ W.
t ≈ 3.2×10⁻¹⁹/(2×10⁻²⁶) = 1.6×10⁷ s, about half a year.
Experimentally emission is essentially instantaneous, contradicting classical gradual energy accumulation.
Final Answer
Classical wave estimate ≈ 1.6×10⁷ s (about 0.5 year), whereas observed emission is immediate.
Exam Tip
This time-lag failure is a central argument for the photon model.
NCERT 11.31Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice.) m_e = 9.11 × 10-31, kg
Given Data
λ = 1 Å = 10⁻¹⁰ m; me = 9.11×10⁻³¹ kg.
Formula Used
X-ray: E = hc/λ. Electron: K = h²/(2mλ²).
Step-by-Step Solution
X-ray energy = (6.626×10⁻³⁴×3×10⁸)/(10⁻¹⁰) = 1.988×10⁻¹⁵ J = 12.4 keV.
Electron energy = (6.626×10⁻³⁴)²/[2(9.11×10⁻³¹)(10⁻¹⁰)²] = 2.41×10⁻¹⁷ J = 150 eV.
Ratio ≈ 12,400/150 ≈ 83.
Final Answer
The 1 Å X-ray photon has about 12.4 keV, while the electron has about 150 eV; the X-ray has roughly 83 times more energy.
Exam Tip
Same wavelength means same momentum, but photon and massive-particle energy relations differ.
NCERT 11.32(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. m_n = 1.675 × 10-27 kg
Given Data
K = 150 eV = 2.403×10⁻¹⁷ J; mn = 1.675×10⁻²⁷ kg.
Formula Used
λ = h/√(2mK).
Step-by-Step Solution
λn = 6.626×10⁻³⁴/√[2(1.675×10⁻²⁷)(2.403×10⁻¹⁷)] = 2.34×10⁻¹² m = 0.0234 Å.
This is much smaller than typical lattice spacing (~1 Å), so the corresponding Bragg angles are extremely small. A 150 eV electron has λ≈1 Å and is much more suitable.
Final Answer
(a) λn = 2.34×10⁻¹² m; (b) no, it is too short for convenient crystal diffraction at ordinary lattice spacings.
Exam Tip
At equal kinetic energy, the heavier particle has the shorter wavelength.
NCERT 11.33An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Given Data
V = 50 kV = 5×10⁴ V; yellow λ≈589 nm.
Formula Used
Non-relativistic λ(Å)=12.27/√V; resolving power ∝1/λ.
Step-by-Step Solution
λe = 12.27/√50000 Å = 0.0549 Å = 5.49×10⁻¹² m. (Relativistic correction gives about 5.36 pm.)
Resolving-power ratio = λyellow/λe = 589×10⁻⁹/(5.49×10⁻¹²) ≈ 1.07×10⁵.
Final Answer
Electron wavelength ≈ 5.5 pm; ideal resolving power is about 10⁵ times that of a yellow-light microscope.
Exam Tip
Actual resolution also depends on lens aberrations and numerical aperture.
NCERT 11.34The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of (10-15) m or less. This structure was first probed in early 1970's using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might be the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV)
Given Data
Required wavelength λ≈10⁻¹⁵ m.
Formula Used
p = h/λ. At ultra-relativistic energy, E≈pc=hc/λ.
Step-by-Step Solution
pc = hc/λ = (6.626×10⁻³⁴×3×10⁸)/(10⁻¹⁵) = 1.988×10⁻¹⁰ J.
In eV, E≈1.988×10⁻¹⁰/(1.602×10⁻¹⁹) = 1.24×10⁹ eV = 1.24 GeV.
This is far above 0.511 MeV, confirming the ultra-relativistic approximation.
Final Answer
Electron-beam energy must be of order 1 GeV or greater.
Exam Tip
For sub-femtometre probes, use relativistic E≈pc, not p²/(2m).
NCERT 11.35Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27°C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.
Given Data
T = 300 K; mHe≈4u=6.64×10⁻²⁷ kg; P=1.013×10⁵ Pa.
Formula Used
λ = h/√(3mkT); number density n=P/(kT); mean separation a≈n⁻¹ᐟ³.
Step-by-Step Solution
λ = 6.626×10⁻³⁴/√[3(6.64×10⁻²⁷)(1.381×10⁻²³)(300)] = 7.3×10⁻¹¹ m.
n = 1.013×10⁵/[1.381×10⁻²³×300] = 2.45×10²⁵ m⁻³.
a = n⁻¹ᐟ³ = 3.44×10⁻⁹ m.
Thus a/λ≈47.
Final Answer
λ≈7.3×10⁻¹¹ m; mean separation≈3.4×10⁻⁹ m, about 47 times larger.
Exam Tip
Because λ is much smaller than separation, helium gas at room temperature behaves nearly classically.
NCERT 11.36Compute the typical de Broglie wavelength of an electron in a metal at 27°C and compare it with the mean separation between two electrons in a metal which is given to be about (2 × 10-10) m.
Given Data
T = 300 K; me = 9.11×10⁻³¹ kg; separation = 2×10⁻¹⁰ m.
Formula Used
Using thermal estimate, λ = h/√(3mekT).
Step-by-Step Solution
λ = 6.626×10⁻³⁴/√[3(9.11×10⁻³¹)(1.381×10⁻²³)(300)] = 6.23×10⁻⁹ m.
Ratio λ/a = 6.23×10⁻⁹/(2×10⁻¹⁰) ≈ 31.
Final Answer
Typical thermal λ≈6.2 nm, about 31 times the stated mean separation.
Exam Tip
Strong overlap of electron matter waves explains why conduction electrons require quantum statistics; a real metal is better described using Fermi energy.
NCERT 11.37Answer the following questions: (a) Quarks inside protons and neutrons are thought to carry fractional charges (+2/3 e ; -1/3 e). Why do they not show up in Millikan's oil-drop experiment? (b) What is so special about the combination (e/m)? Why do we not simply talk of (e) and (m) separately? (c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures? (d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons? (e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: E = hν,; p = h/λ But while the value of (λ) is physically significant, the value of (ν) (and therefore, the value of the phase speed (vλ)) has no physical significance. Why?
Given Data
Conceptual question combining confinement, charged-particle dynamics, gas discharge, photoelectric emission and matter waves.
Formula Used
Use quark confinement; F=qE and a=qE/m; mean free path; Einstein's photoelectric equation; group velocity of a wave packet.
Step-by-Step Solution
(a) Quarks are confined inside hadrons by the strong interaction and are not observed as isolated free particles; oil drops therefore acquire charge only in integral multiples of e.
(b) In electric or magnetic deflection, acceleration or curvature depends on charge-to-mass ratio: a=(e/m)E and r=mv/(eB). The experiment therefore determines e/m directly; separate e and m require an additional independent measurement.
(c) At ordinary pressure the mean free path is small, so electrons lose energy in frequent collisions before they can ionize molecules. At reduced pressure they gain sufficient energy between collisions to cause ionization and an avalanche discharge. At extremely low pressure there may again be too few gas molecules to sustain conduction.
(d) Electrons originate at different depths and energy states and lose different amounts of energy while reaching the surface. Only surface electrons with favourable initial energies emerge with Kmax=hν−Φ; others have lower energies.
(e) Wavelength determines observable interference and diffraction. A free-particle matter wave is represented by a wave packet whose group velocity equals particle velocity. Its phase frequency and phase velocity are not directly observable particle-motion quantities; phase velocity can exceed c without carrying information.
Final Answer
(a) Quark confinement; (b) trajectories measure e/m; (c) low pressure increases mean free path enough for ionization; (d) electrons suffer unequal binding and energy losses; (e) wavelength and group velocity are observable, while phase frequency/speed have no direct standalone particle interpretation.
Exam Tip
For conceptual subparts, connect each answer to one governing physical principle rather than memorizing isolated statements.
05
NEET practice
NEET Dual Nature Questions
25 fresh, non-duplicate MCQs covering photon energy, work function, threshold, intensity, matter waves and Davisson-Germer.
Q1Light of wavelength 400 nm falls on a metal of work function 2.0 eV. The stopping potential is closest to
Correct answer: B
Explanation: Photon energy = 1240/400 = 3.10 eV. Kmax = 3.10 - 2.0 = 1.10 eV, so V0 = 1.10 V.
Q2For a given photocathode, doubling the intensity of light above threshold mainly doubles the
Correct answer: C
Explanation: Intensity increases photon number per second, so more electrons are emitted. It does not change the energy per photon.
Q3A metal has threshold wavelength 620 nm. Radiation of wavelength 700 nm is incident with very high intensity. The result is
Correct answer: C
Explanation: 700 nm is longer than threshold wavelength, so photon energy is below the work function. Intensity cannot compensate.
Q4If frequency is increased while intensity is kept fixed in a photoelectric experiment, the stopping potential
Correct answer: B
Explanation: V0 = (h/e)ν - Φ/e, so V0 rises linearly with frequency above threshold.
Q5The slope of a stopping-potential versus frequency graph is
Correct answer: C
Explanation: From eV0 = hν - Φ, V0 = (h/e)ν - Φ/e.
Q6The de Broglie wavelength of an electron accelerated through voltage V is proportional to
Correct answer: D
Explanation: λ = h/√(2meV), hence λ ∝ V-1/2.
Q7An electron wavelength at 100 V is about
Correct answer: C
Explanation: λ = 12.27/√100 = 1.227 Å.
Q8In Davisson-Germer experiment, a sharp peak in scattered electron intensity proves
Correct answer: B
Explanation: Diffraction maxima arise from constructive wave interference.
Q9If the work function of a metal is 3.1 eV, the threshold frequency is approximately
Correct answer: A
Explanation: ν0 = Φ/h. Since h = 4.136×10-15 eV s, ν0 = 3.1/4.136×1014 = 7.5×1014 Hz.
Q10Photoelectric emission is immediate because
Correct answer: B
Explanation: Einstein's one-photon one-electron interaction explains the absence of classical time lag.
Q11A photon and an electron have the same wavelength. They necessarily have the same
Correct answer: C
Explanation: p = h/λ for both. Their energy relations are different.
Q12If a light beam has photon energy below work function, increasing intensity causes
Correct answer: D
Explanation: Every photon is still individually unable to liberate an electron.
Q13The maximum kinetic energy of photoelectrons is measured by
Correct answer: B
Explanation: Kmax = eV0.
Q14The threshold wavelength of a metal of work function 2.5 eV is
Correct answer: B
Explanation: λ0 = 1240/2.5 = 496 nm.
Q15In a photoelectric current versus anode potential graph, saturation current increases when
Correct answer: B
Explanation: More photons per second produce more emitted electrons per second.
Q16For a fixed accelerating voltage, a proton has a shorter de Broglie wavelength than an electron because the proton has
Correct answer: C
Explanation: For same V and charge magnitude, λ ∝ 1/√m.
Q17Bragg diffraction condition is written as
Correct answer: B
Explanation: Constructive interference from crystal planes satisfies nλ = 2d sinθ.
Q18The unit of h/e, the slope of V0 versus ν, is
Correct answer: A
Explanation: V0/ν has units V/Hz = V s.
Q19A 300 nm photon has energy closest to
Correct answer: C
Explanation: E = 1240/300 = 4.13 eV.
Q20Matter waves are difficult to observe for a moving cricket ball because
Correct answer: B
Explanation: λ = h/p becomes unimaginably small for macroscopic momentum.
Q21A metal has Φ = 4.0 eV. Which light can emit photoelectrons?
Correct answer: D
Explanation: Required E ≥ 4 eV. Only 250 nm gives E = 4.96 eV.
Q22The photoelectric effect supports particle nature of light mainly because
Correct answer: A
Explanation: A frequency threshold cannot be explained by classical wave intensity alone.
Q23If electron accelerating voltage changes from 54 V to 216 V, its wavelength becomes
Correct answer: C
Explanation: Voltage is multiplied by 4, so λ is divided by 2.
Q24In V0 versus ν graph, the frequency-axis intercept gives
Correct answer: B
Explanation: At V0 = 0, photon energy just equals work function, so ν = ν0.
Q25Electron diffraction is strongest when electron wavelength is comparable with
Correct answer: B
Explanation: Diffraction needs wavelength comparable with the spacing that creates path difference.
06
JEE Main
JEE Main Dual Nature Questions
25 numerical and conceptual questions with key formula, step-by-step solution and final answer.
M1A metal of work function 2.25 eV is illuminated by 360 nm light. Find stopping potential.
Key formula: V0 = 1240/λ(nm) - Φ(eV)
Solution: Photon energy = 1240/360 = 3.444 eV. Kmax = 3.444 - 2.25 = 1.194 eV. Therefore V0 = 1.19 V.
Final answer: 1.19 V
M2A photocathode has threshold frequency 5.0×1014 Hz. Light of frequency 8.0×1014 Hz is used. Find V0.
Key formula: V0 = (h/e)(ν - ν0)
Solution: h/e = 4.136×10-15 V s. V0 = 4.136×10-15 × 3.0×1014 = 1.24 V.
Final answer: 1.24 V
M3Light of wavelength 500 nm gives stopping potential 0.40 V. Find work function.
Key formula: Φ(eV) = 1240/λ(nm) - V0
Solution: Photon energy = 1240/500 = 2.48 eV. Work function = 2.48 - 0.40 = 2.08 eV.
Final answer: 2.08 eV
M4The slope of V0 versus frequency graph is 4.10×10-15 V s. Estimate h.
Key formula: slope = h/e
Solution: h = e × slope = 1.602×10-19 × 4.10×10-15 = 6.57×10-34 J s.
Final answer: 6.57×10-34 J s
M5An electron beam is accelerated through 150 V. Find de Broglie wavelength in Å.
Key formula: λ(Å) = 12.27/√V
Solution: λ = 12.27/√150 = 12.27/12.247 = 1.00 Å.
Final answer: 1.00 Å
M6Find the accelerating voltage required for electron wavelength 0.20 nm.
Key formula: λ(Å) = 12.27/√V
Solution: 0.20 nm = 2.0 Å. V = (12.27/2.0)2 = 37.6 V.
Final answer: 37.6 V
M7Compare de Broglie wavelengths of electron and proton accelerated through the same potential difference.
Key formula: λ = h/√(2mqV)
Solution: Both have charge magnitude e, so λ ∝ 1/√m. λp/λe = √(me/mp) ≈ 1/√1836 = 1/42.8.
Final answer: Proton wavelength is about 1/43 of electron wavelength.
M8For d = 1.0 Å, first-order Bragg maximum occurs at θ = 30°. Find λ.
Key formula: nλ = 2d sinθ
Solution: λ = 2(1.0)sin30° = 1.0 Å.
Final answer: 1.0 Å
M9An electron has de Broglie wavelength 1.5 Å. Find its kinetic energy in eV.
Key formula: λ(Å)=12.27/√V and K=eV
Solution: V = (12.27/1.5)2 = 66.9 V. Electron kinetic energy = 66.9 eV.
Final answer: 66.9 eV
M10Find photon momentum for light of wavelength 600 nm.
Key formula: p = h/λ
Solution: p = 6.626×10-34/(600×10-9) = 1.10×10-27 kg m s-1.
Final answer: 1.10×10-27 kg m s-1
M11A 2 mW laser emits 500 nm photons. Estimate photons emitted per second.
Key formula: N = P/(hc/λ)
Solution: Photon energy = 1240/500 = 2.48 eV = 3.97×10-19 J. N = 2×10-3/3.97×10-19 = 5.0×1015 s-1.
Final answer: 5.0×1015 photons s-1
M12A surface emits for 450 nm but not for 600 nm. Which interval contains threshold wavelength?
Key formula: Emission requires λ ≤ λ0
Solution: Since 450 nm emits, λ0 ≥ 450 nm. Since 600 nm does not emit, λ0 < 600 nm.
Final answer: 450 nm ≤ λ0 < 600 nm
M13Stopping potentials for frequencies 6×1014 Hz and 8×1014 Hz differ by how much?
Key formula: ΔV0 = (h/e)Δν
Solution: ΔV0 = 4.136×10-15 × 2×1014 = 0.827 V.
Final answer: 0.827 V
M14A photoelectron has Kmax = 0.80 eV. Find maximum speed.
Key formula: K = ½mv2
Solution: K = 0.80×1.602×10-19 J. v = √(2K/me) = 5.30×105 m s-1.
Final answer: 5.30×105 m s-1
M15For a metal of Φ = 2.7 eV, calculate threshold wavelength.
Key formula: λ0(nm) = 1240/Φ(eV)
Solution: λ0 = 1240/2.7 = 459 nm.
Final answer: 459 nm
M16A graph of Kmax in eV versus frequency cuts the frequency axis at 4×1014 Hz. Find work function.
Key formula: Φ = hν0
Solution: Φ(eV) = 4.136×10-15 × 4×1014 = 1.65 eV.
Final answer: 1.65 eV
M17A beam current is 3.2 μA. How many electrons cross a section per second?
Key formula: N = I/e
Solution: N = 3.2×10-6/1.602×10-19 = 2.0×1013 electrons s-1.
Final answer: 2.0×1013 s-1
M18A neutron of kinetic energy 0.025 eV has what order of wavelength?
Key formula: λ = h/√(2mK)
Solution: Using mn = 1.675×10-27 kg and K = 0.025 eV gives λ ≈ 1.8 Å, comparable with crystal spacing.
Final answer: About 1.8 Å
M19What voltage change is needed to reduce electron wavelength by a factor of 3?
Key formula: λ ∝ 1/√V
Solution: λ2 = λ1/3 requires √V2 = 3√V1, hence V2 = 9V1.
Final answer: Voltage must be made 9 times.
M20An X-ray tube operates at 20 kV. Find minimum wavelength.
Key formula: λmin(nm) = 1240/V(eV)
Solution: Electron energy = 20,000 eV. λmin = 1240/20000 = 0.062 nm = 0.62 Å.
Final answer: 0.062 nm
M21A 250 nm photon falls on a 4.5 eV metal. Find Kmax.
Key formula: Kmax = 1240/λ - Φ
Solution: Photon energy = 1240/250 = 4.96 eV. Kmax = 4.96 - 4.5 = 0.46 eV.
Final answer: 0.46 eV
M22For the same electron beam, if Bragg angle increases, what happens to interplanar spacing for first order?
Key formula: d = λ/(2sinθ)
Solution: With λ fixed, d is inversely proportional to sinθ. Larger θ means larger sinθ and smaller d.
Final answer: Interplanar spacing is smaller.
M23An electron and photon have equal momentum p. Compare their wavelengths.
Key formula: λ = h/p
Solution: Since both follow p = h/λ, equal momentum means equal wavelength.
Final answer: They have equal wavelengths.
M24A photocell has saturation current 10 μA. If light intensity is reduced to 40 percent, estimate new saturation current.
Key formula: Is ∝ intensity
Solution: New saturation current = 0.40 × 10 μA = 4 μA, assuming frequency remains above threshold.
Final answer: 4 μA
M25Electron microscope uses 10 kV electrons. Estimate wavelength.
Key formula: λ(Å)=12.27/√V
Solution: λ = 12.27/√10000 = 0.1227 Å = 0.01227 nm.
Final answer: 0.1227 Å
07
JEE Advanced
JEE Advanced Difficult Questions on Dual Nature
15 concept-rich problems including graph-based, multiple-correct, assertion-reason, matching and paragraph styles.
A1For two metals A and B, V0 versus ν lines are parallel, but A cuts the frequency axis at a lower frequency. Which statements are correct?
Multiple-correct
Solution: Parallel lines mean same slope h/e, as expected for all metals. The x-intercept is ν0 = Φ/h. Lower intercept means smaller work function. At same ν, V0 = (h/e)(ν - ν0), so smaller ν0 gives larger V0.
Final answer: Correct: same Planck constant slope, ΦA < ΦB, and at a fixed high frequency A has larger stopping potential.
A2A metal gives stopping potentials 0.5 V and 1.7 V for frequencies 6×1014 Hz and 9×1014 Hz. Find h/e and threshold frequency.
Graph based
Solution: Slope = ΔV/Δν = 1.2/(3×1014) = 4.0×10-15 V s. Use V0 = slope(ν - ν0). With V0=0.5 V at 6×1014 Hz, ν0 = 6×1014 - 0.5/(4×10-15) = 4.75×1014 Hz.
Final answer: h/e = 4.0×10-15 V s; ν0 = 4.75×1014 Hz.
A3A 300 nm photon ejects an electron from a 2.4 eV metal. The electron then enters a region where it is accelerated through 100 V. Find its final de Broglie wavelength.
Multi-concept numerical
Solution: Initial maximum kinetic energy = 1240/300 - 2.4 = 1.73 eV. After additional acceleration through 100 V, K = 101.73 eV. For electron λ = 12.27/√101.73 = 1.216 Å.
Final answer: Approximately 1.21 Å.
A4In a Davisson-Germer setup, the angle between incident and scattered beams is 50°. A crystal-plane geometry makes Bragg angle θ = 65°. For d = 0.91 Å and n = 1, calculate λ and comment on using 50° directly.
Angle convention
Solution: Bragg law needs the glancing angle with the crystal plane. λ = 2d sinθ = 2(0.91)sin65° = 1.65 Å. The scattering angle is not automatically θ; the diagram's geometry must be read.
Final answer: λ = 1.65 Å; using 50° directly is a geometry error.
A5At fixed frequency above threshold, intensity is doubled. Which quantities may double: photon flux, saturation current, stopping potential, maximum kinetic energy?
Multiple-correct
Solution: Doubling intensity doubles the number of photons per second and usually the number of emitted electrons per second before space-charge limits. Photon energy hν is unchanged, so Kmax and V0 are unchanged.
Final answer: Photon flux and saturation current may double.
A6A metal is exposed to 450 nm light and emits electrons with V0 = 0.30 V. The source is replaced by 300 nm light of the same power. Find the new stopping potential and explain why current need not be the same.
Paragraph
Solution: Work function = 1240/450 - 0.30 = 2.456 eV. For 300 nm, photon energy = 4.133 eV, so V0 = 4.133 - 2.456 = 1.677 V. Same optical power at higher photon energy means fewer photons per second, so current is not guaranteed to be equal.
Final answer: V0 ≈ 1.68 V; current may differ because photon rate changes.
A7Match: (A) Kmax-ν slope, (B) V0-ν slope, (C) λe-V dependence, (D) Bragg graph n versus sinθ at fixed λ,d.
Matching type
Solution: Kmax = hν - Φ gives slope h. V0 = (h/e)ν - Φ/e gives slope h/e. Electron wavelength varies as 1/√V. From nλ = 2d sinθ, n is proportional to sinθ.
Final answer: A-h, B-h/e, C-V-1/2, D-linear.
A8Assertion: A high-intensity red beam below threshold cannot produce photoelectrons. Reason: Work function is energy needed per electron, and each photon must individually supply at least this much energy.
Assertion-reason
Solution: Photoelectric emission is not caused by slow accumulation of classical wave energy. In the photon model one photon interacts with one electron. If hν < Φ, emission does not occur even at high intensity.
Final answer: Both are true and the reason explains the assertion.
A9Electron wavelength changes from 1.8 Å to 1.2 Å when accelerating voltage is changed from V1 to V2. Find V1, V2, and V2/V1.
Difficult numerical
Solution: V = (12.27/λ)2. So V1 = (12.27/1.8)2 = 46.5 V and V2 = (12.27/1.2)2 = 104.6 V. Ratio = (1.8/1.2)2 = 2.25.
Final answer: V1 = 46.5 V, V2 = 104.6 V, ratio = 2.25.
A10An electron and alpha particle are accelerated through the same potential V. Which statements are correct?
Multiple-correct
Solution: K = qV, so alpha with charge 2e has 2eV while electron has eV. Momentum p = √(2mqV) is much larger for alpha due to larger m and q. Therefore λ = h/p is shorter.
Final answer: Alpha has larger kinetic energy, larger momentum, and shorter wavelength.
A11For a metal, photons of 4 eV and 5 eV give stopping potentials 0.6 V and 1.6 V. Find work function and test consistency.
Data interpretation
Solution: Using Kmax(eV)=V0, Φ = E - V0. From first, Φ = 4 - 0.6 = 3.4 eV. From second, Φ = 5 - 1.6 = 3.4 eV. Same result means consistency.
Final answer: Φ = 3.4 eV, data are consistent.
A12For d = 1.2 Å and λ = 1.0 Å, list possible Bragg orders and corresponding θ values.
Bragg order
Solution: sinθ = nλ/(2d) = n/2.4. For n=1, sinθ=0.4167 and θ=24.6°. For n=2, sinθ=0.8333 and θ=56.4°. For n=3, sinθ=1.25, impossible.
Final answer: n = 1 gives θ = 24.6°, n = 2 gives θ = 56.4°; n ≥ 3 impossible.
A13A photon and an electron each have wavelength 1 Å. Compare their energies and explain the difference.
Concept-rich
Solution: Same λ means same momentum p=h/λ. Photon energy is pc = hc/λ = 12.4 keV. Non-relativistic electron kinetic energy is p2/2m = h2/(2mλ2) ≈ 150 eV. Equal momentum does not imply equal energy.
Final answer: Photon energy = 12.4 keV; electron kinetic energy ≈ 150 eV.
A14A photocell has work function 2.0 eV. It is illuminated by two lights: 620 nm and 310 nm, same intensity. Discuss emission, stopping potential, and saturation current trend.
Paragraph
Solution: 620 nm photons have energy 1240/620 = 2.0 eV, so electrons barely escape and V0 ≈ 0. 310 nm photons have energy 4.0 eV, so Kmax ≈ 2.0 eV and V0 ≈ 2.0 V. Same intensity does not mean same photon rate: higher-energy photons are fewer per second for the same power.
Final answer: 620 nm is threshold-like with nearly zero K; 310 nm gives about 2.0 V stopping potential; currents depend on photon rate and quantum efficiency.
A15Show how de Broglie condition gives Bohr angular momentum quantization.
Bohr link
Solution: For a standing matter wave around a circular orbit, circumference must contain an integer number of wavelengths: 2πr = nλ. Substitute λ = h/mv to get 2πr = nh/mv, hence mvr = nh/2π.
Final answer: mvr = nh/2π.
08
IB Physics
IB Physics Structured Questions
Explanation, data analysis, graph interpretation and experimental reasoning for international learners.
Q1Explain why photoelectric emission supports a photon model.
Answer: A threshold frequency and instantaneous emission are naturally explained if light energy arrives in photons of energy hν. Classical wave intensity alone cannot explain why below-threshold light fails.
Q2A graph of stopping potential against frequency is linear. State what the gradient and intercept represent.
Answer: Gradient is h/e. The frequency-axis intercept is threshold frequency, and the voltage-axis intercept is -Φ/e.
Q3Describe one uncertainty in measuring stopping potential.
Answer: The current may approach zero gradually because emitted electrons have a distribution of energies, so the zero-current point can be uncertain.
Q4Why does increasing intensity increase photocurrent but not stopping potential?
Answer: Intensity changes photon flux. Stopping potential depends on the maximum energy per emitted electron, set by photon frequency.
Q5A 405 nm source is used on a metal of Φ = 2.1 eV. Calculate Kmax.
Answer: E = 1240/405 = 3.06 eV. Kmax = 0.96 eV.
Q6How does Davisson-Germer evidence combine particle and wave ideas?
Answer: Electrons are detected as particles through current, but their angular distribution has diffraction maxima predicted by wave interference.
Q7State why vacuum is required in an electron diffraction experiment.
Answer: Vacuum reduces collisions with gas molecules that would scatter electrons and change their energy.
Q8In data analysis, why should a best-fit line be used for V0 versus ν?
Answer: Experimental points include uncertainty. A best-fit line gives a more reliable value of h/e than one pair of points.
Q9Explain the significance of electron wavelength being comparable with crystal spacing.
Answer: Comparable length scales allow path differences of order λ, producing observable constructive and destructive interference.
Q10Suggest a reason why measured photocurrent may not double when intensity doubles.
Answer: Space charge, collection efficiency, surface condition, and quantum efficiency can prevent exact proportionality.
09
ICSE and IGCSE
ICSE / IGCSE Dual Nature Questions
15 school-level questions with short, conceptually strong answers.
Q1What is meant by work function?
Answer: The minimum energy required to remove an electron from the surface of a metal.
Q2Define threshold frequency.
Answer: The minimum frequency of incident radiation needed for photoelectric emission.
Q3Why can blue light eject electrons when red light cannot for some metals?
Answer: Blue light has higher frequency and higher photon energy.
Q4What is stopping potential?
Answer: The reverse potential that just stops the fastest emitted photoelectrons from reaching the collector.
Q5State the formula for photon energy.
Answer: E = hν = hc/λ.
Q6Calculate photon energy for 620 nm light in eV.
Answer: E = 1240/620 = 2.0 eV.
Q7What happens to photoelectric current when intensity increases above threshold?
Answer: It increases because more photons strike the surface per second.
Q8Does increasing intensity change threshold frequency?
Answer: No. Threshold frequency depends on the metal.
Q9State de Broglie's relation.
Answer: λ = h/p.
Q10Why is electron diffraction not seen for slow everyday objects?
Answer: Everyday objects have large momentum, so their de Broglie wavelength is extremely small.
Q11Name the experiment that verified wave nature of electrons.
Answer: Davisson-Germer experiment.
Q12What is Bragg's law?
Answer: nλ = 2d sinθ.
Q13Why is a crystal used in electron diffraction?
Answer: Its regular atomic planes act like a diffraction grating.
Q14If voltage accelerating electrons increases, what happens to wavelength?
Answer: It decreases as λ ∝ 1/√V.
Q15What does a bright ring in an electron diffraction pattern represent?
Answer: A direction or angle where scattered electron waves interfere constructively.
10
A-Level and AP
A-Level / AP Physics Practice
Photon model, wave-particle duality, electron diffraction, graph analysis and experimental interpretation.
1AP Physics A 500 nm photon strikes a 2.0 eV metal. Find maximum kinetic energy.
Answer: E = 1240/500 = 2.48 eV, so Kmax = 0.48 eV.
2AP Physics Sketch how stopping potential changes with frequency.
Answer: It is zero below threshold in practice, then increases linearly with slope h/e above threshold.
3A-Level Derive λ = h/√(2meV) for an electron.
Answer: Use eV = ½mv2, so p = mv = √(2meV). Substitute into λ = h/p.
4A-Level Why do electrons accelerated through tens of volts diffract from crystals?
Answer: Their wavelengths are of order angstroms, comparable with crystal plane spacing.
5AP Physics If electron voltage is quadrupled, what happens to diffraction-ring radius qualitatively?
Answer: Wavelength halves. For small angles, diffraction angles and ring radii decrease.
6A-Level Explain why photon momentum is p = h/λ.
Answer: For photons E = hf and E = pc. Since c = fλ, p = hf/c = h/λ.
7AP Physics A source emits 3 eV photons with power 6 mW. Find photon rate.
Answer: N = P/E = 6×10-3/(3×1.602×10-19) = 1.25×1016 s-1.
8A-Level State one reason electron microscopes have high resolving power.
Answer: Electrons can have wavelengths much shorter than visible light when accelerated through large voltages.
9AP Physics A metal emits under 350 nm light but not 450 nm light. Bound its work function.
Answer: 1240/450 < Φ ≤ 1240/350, so 2.76 eV < Φ ≤ 3.54 eV.
10A-Level Describe what changes in Davisson-Germer data when accelerating voltage increases.
Answer: Electron wavelength decreases, so the constructive-interference angle changes according to Bragg's law.
11
Case based
Case-Based Questions
Five exam-style case studies with subquestions and direct answers.
Case Study 1
Photoelectric experiment with two wavelengths
A metal of work function 2.2 eV is illuminated first by 500 nm light and then by 300 nm light at the same power.
1. Does 500 nm light emit photoelectrons?
Yes. E = 1240/500 = 2.48 eV, slightly above the work function.
2. Find V0 for 500 nm.
Kmax = 2.48 - 2.2 = 0.28 eV, so V0 = 0.28 V.
3. Find V0 for 300 nm.
E = 4.13 eV, so V0 = 4.13 - 2.2 = 1.93 V.
4. Which source has larger photon rate at the same power?
500 nm light, because each photon has less energy.
Case Study 2
Stopping-potential graph
A line of V0 versus frequency has slope 4.14×10-15 V s and cuts the frequency axis at 5.5×1014 Hz.
1. What is Planck's constant from the slope?
h = e × slope = 6.63×10-34 J s.
2. Find the work function in eV.
Φ = hν0 = 4.14×10-15 × 5.5×1014 = 2.28 eV.
3. Find threshold wavelength.
λ0 = c/ν0 = 545 nm.
4. What is V0 at 8.0×1014 Hz?
V0 = 4.14×10-15(2.5×1014) = 1.04 V.
Case Study 3
Davisson-Germer nickel peak
A 54 V electron beam gives a strong scattered intensity maximum from nickel. The effective Bragg angle is 65° and d = 0.91 Å.
1. Find de Broglie wavelength from voltage.
λ = 12.27/√54 = 1.67 Å.
2. Find wavelength from Bragg law.
λ = 2(0.91)sin65° = 1.65 Å.
3. What does the agreement show?
Electron beams have matter waves obeying λ = h/p.
4. Why must the chamber be evacuated?
To prevent gas collisions from scattering and slowing electrons.
Case Study 4
Electron accelerated through voltage
An electron is accelerated from rest through 25 V, then through 100 V in a second run.
1. Find λ at 25 V.
λ = 12.27/5 = 2.454 Å.
2. Find λ at 100 V.
λ = 12.27/10 = 1.227 Å.
3. How does momentum change?
Momentum doubles because p ∝ √V.
4. How does diffraction angle change for fixed crystal planes?
It decreases because sinθ ∝ λ.
Case Study 5
Bragg diffraction decision
Electrons of wavelength 0.90 Å strike a crystal with d = 1.00 Å.
1. Is first order possible?
Yes. sinθ = 0.90/2 = 0.45, so θ = 26.7°.
2. Is second order possible?
Yes. sinθ = 1.80/2 = 0.90, so θ = 64.2°.
3. Is third order possible?
No. sinθ would be 2.70/2 = 1.35, impossible.
4. What common mistake is avoided here?
Checking sinθ ≤ 1 before accepting an order.
12
Error control
Common Mistakes Students Make in Dual Nature
The mistakes that most often cost marks in boards, NEET, JEE and international exams.
Intensity versus frequency
Intensity controls photon number and photocurrent; frequency controls energy per photon and stopping potential.
nm conversion
Use E(eV) = 1240/λ(nm) only when λ is in nm, not metres or angstroms.
Stopping potential sign
Use the magnitude for energy: Kmax = eV0. The applied potential is retarding.
Threshold myth
High intensity below threshold cannot cause emission in the ordinary photoelectric effect.
Wrong graph slope
Kmax versus ν has slope h; V0 versus ν has slope h/e.
Photon vs electron energy
A photon and electron with the same wavelength have the same momentum, not the same energy.
Electron shortcut unit
λ = 12.27/√V gives λ in Å only when V is in volts.
Bragg angle confusion
Do not insert the scattering angle directly unless the diagram says it equals the Bragg angle.
Using all emitted electrons
Einstein's equation gives maximum kinetic energy. Other electrons may emerge slower.
Macroscopic matter waves
All moving bodies have λ = h/p, but large p makes ordinary wavelengths unobservable.
13
Teaching approach
How Kumar Sir Explains This Chapter
Kumar Sir explains Dual Nature concept-first, not by rote learning.
Concept-first Physics coaching
Kumar Sir uses diagram-based explanation, numerical practice, graph interpretation and focused doubt clearing so students understand why each formula works before using it in exam problems.
- 30+ years of teaching experience
- NEET, JEE Main, JEE Advanced and CBSE expertise
- IB, ICSE, IGCSE, AP Physics and A-Level support
- One-on-one online Physics classes
- Photoelectric graphs, de Broglie numericals and Davisson-Germer geometry explained step by step
Contact Kumar Sir
Phone / WhatsApp: +91 9958461445
Email: kumarsirphysics@gmail.com
Website: https://kumarphysicsclasses.com
14
Final revision
Final Revision for Dual Nature
One-page formula box, graph summary, top 20 exam facts and last-minute checklist.
One-Page Formula Box
E = hν = hc/λ
Kmax = hν - Φ = eV0
ν0 = Φ/h, λ0 = hc/Φ
λ = h/p = h/mv = h/√(2mK)
λe(Å) = 12.27/√V, nλ = 2d sinθ
Graph Summary
- Photoelectric current versus intensity: straight-line increase above threshold.
- Current versus anode potential: saturation at positive potential and zero current at stopping potential.
- V0 versus ν: straight line with slope h/e and intercept ν0.
- Kmax versus ν: straight line with slope h.
- Electron wavelength versus voltage: decreasing curve proportional to V-1/2.
Last-Minute Checklist
- Check wavelength units before using 1240/λ.
- Use maximum kinetic energy with stopping potential.
- Separate intensity effects from frequency effects.
- Confirm Bragg angle from diagram geometry.
- Write final numerical answers with units.
Top 20 Exam Facts
- 1Photon energy is proportional to frequency and inversely proportional to wavelength.
- 2Work function is a surface property of the metal.
- 3Threshold frequency depends on the material, not on intensity.
- 4Below threshold there is no photoelectric emission in the single-photon model.
- 5Stopping potential measures maximum kinetic energy.
- 6Saturation current depends mainly on intensity above threshold.
- 7V0 versus ν slope is h/e.
- 8Kmax versus ν slope is h.
- 9Frequency-axis intercept of V0 graph is ν0.
- 10Photon momentum is h/λ.
- 11de Broglie wavelength is h/p.
- 12For an electron accelerated through V, λ ∝ 1/√V.
- 13Electron shortcut λ(Å)=12.27/√V is non-relativistic.
- 14Davisson-Germer verified electron matter waves.
- 15Crystal planes produce constructive interference at Bragg angles.
- 16Bragg law uses nλ = 2d sinθ.
- 17Diffraction is strong when wavelength is comparable with spacing.
- 18Macroscopic matter waves are too small to detect easily.
- 19Same wavelength means same momentum for photon and particle.
- 20Same optical power at different frequencies means different photon rates.