NCERT Solutions for Class 12 Physics Chapter 11: Dual Nature of Radiation and Matter

Concept and context. X-rays are produced when fast electrons are suddenly stopped on striking a metal target: the kinetic energy of the electrons is converted into radiation (bremsstrahlung). This is the inverse of the photoelectric effect — an electron of kinetic energy K can create a photon of energy hν at most equal to K. An electron accelerated through a potential difference V acquires a kinetic energy eV. In the most energetic conversion all the kinetic energy of the electron is transferred to a single photon, giving the maximum frequency (the Duane–Hunt limit): hν_max = eV and, correspondingly, the minimum wavelength λ_min = c/ν_max = hc/(eV).

Given: V = 30 kV = 3.0 × 10⁴ V; e = 1.6 × 10⁻¹⁹ C; h = 6.63 × 10⁻³⁴ J s; c = 3.0 × 10⁸ m s⁻¹.

(a) Maximum frequency

1. Energy of the electrons: K = eV = 1.6 × 10⁻¹⁹ × 3.0 × 10⁴ = 4.8 × 10⁻¹⁵ J (= 30 keV).
2. hν_max = eV ⇒ ν_max = eV/h = (4.8 × 10⁻¹⁵)/(6.63 × 10⁻³⁴) = 7.24 × 10¹⁸ Hz.

(b) Minimum wavelength

1. λ_min = c/ν_max = (3.0 × 10⁸)/(7.24 × 10¹⁸) = 4.14 × 10⁻¹¹ m.
2. λ_min = 4.1 × 10⁻¹¹ m = 0.041 nm (= 0.41 Å). (Handy formula: λ_min (nm) = 1.24/V(kV) = 1.24/30 = 0.0413 nm.)

ν_max = eV/h = 7.2 × 10¹⁸ Hz ; λ_min = hc/eV = 4.1 × 10⁻¹¹ m

• Interpretation: the X-ray spectrum from such a tube is continuous but has a sharp cut-off at λ_min, which depends only on the applied voltage (not on the target material); raising the voltage shortens λ_min (higher penetrating power).
• Region of the spectrum: 0.041 nm lies in the X-ray (and the beginning of the γ-ray) region — NCERT lists X-rays from about 10 nm to 10⁻⁴ nm.

Conclusion: the maximum frequency of the X-rays is 7.2 × 10¹⁸ Hz and the minimum wavelength is 4.1 × 10⁻¹¹ m (0.041 nm).

Concept and context. In Einstein’s photoelectric equation (NCERT Section 11.6) a single photon of energy hν gives all its energy to a single electron; part of it (the work function φ₀) is used to free the electron from the metal and the remainder appears as the electron’s kinetic energy: K_max = hν − φ₀ (more tightly bound electrons emerge with smaller energies). The stopping potential V₀ is the retarding potential at which even the fastest photoelectrons are stopped: eV₀ = K_max (NCERT Fig. 11.3, 11.5). Photoemission is possible only if hν > φ₀, i.e. ν > ν₀ = φ₀/h.

Given: φ₀ = 2.14 eV; ν = 6 × 10¹⁴ Hz; h = 6.63 × 10⁻³⁴ J s; 1 eV = 1.6 × 10⁻¹⁹ J; m = 9.11 × 10⁻³¹ kg.

(a) Maximum kinetic energy

1. Photon energy: hν = (6.63 × 10⁻³⁴)(6 × 10¹⁴) = 3.98 × 10⁻¹⁹ J = 3.98 × 10⁻¹⁹/1.6 × 10⁻¹⁹ = 2.49 eV.
2. K_max = hν − φ₀ = 2.49 − 2.14 = 0.35 eV (= 5.5 × 10⁻²⁰ J).

(b) Stopping potential

eV₀ = K_max ⇒ V₀ = 0.35 eV/e = 0.35 V

(c) Maximum speed of the photoelectrons

1. K_max = ½ m v²_max ⇒ v_max = √(2K_max/m).
2. v_max = √[2 × 5.5 × 10⁻²⁰/(9.11 × 10⁻³¹)] = √(1.2 × 10¹¹) = 3.5 × 10⁵ m s⁻¹.

• Check: ν = 6.0 × 10¹⁴ Hz is greater than ν₀ = 5.16 × 10¹⁴ Hz, so photoemission occurs; the corresponding wavelength is 500 nm (green light). The speed is a small fraction of c (10⁻³), so the non-relativistic formula is valid.
• Intensity would only change the number of photoelectrons per second (the photocurrent), not K_max.

Conclusion: (a) K_max = 0.35 eV; (b) stopping potential = 0.35 V; (c) maximum speed = 3.5 × 10⁵ m s⁻¹.

Concept and context. In the photoelectric experiment (NCERT Fig. 11.1), when the collector plate A is made negative with respect to the emitter C, the electrons are decelerated. At the stopping (cut-off) potential V₀ the photocurrent just becomes zero: even the most energetic electrons (emitted with K_max) are brought to rest just before reaching the collector. The work done by the retarding field, eV₀, equals the maximum kinetic energy: K_max = eV₀. This gives a direct way of measuring the maximum energy of photoelectrons (and, by Einstein’s equation, h/e).

• Given: V₀ = 1.5 V.
• K_max = eV₀ = e × 1.5 V = 1.5 eV = 1.5 × 1.6 × 10⁻¹⁹ J = 2.4 × 10⁻¹⁹ J.

K_max = eV₀ = 1.5 eV = 2.4 × 10⁻¹⁹ J

• Remark: when the energy is measured in electron-volts it is numerically equal to the stopping potential in volts — the great convenience of the eV unit.
• Independent of intensity: V₀ (hence K_max) does not change when the intensity of the light is increased; only the saturation current does (NCERT Fig. 11.3). It rises with the frequency (Fig. 11.4).
• Possible follow-up: with Einstein’s equation hν = φ₀ + eV₀ the work function of the emitter can be found if the frequency is known (Exercise 11.9).

Conclusion: the maximum kinetic energy of the photoelectrons is 1.5 eV (2.4 × 10⁻¹⁹ J).

Concept and context. In the photon picture of light (NCERT Section 11.7) a beam of monochromatic light of wavelength λ (frequency ν = c/λ) consists of photons, each carrying energy E = hν = hc/λ and momentum p = h/λ = E/c, independent of the intensity of the beam. The intensity (power) of the beam is determined by the number of photons per second N: P = N E. Photons are electrically neutral and always move with the speed c.

Given: λ = 632.8 nm = 6.328 × 10⁻⁷ m; P = 9.42 mW = 9.42 × 10⁻³ W; h = 6.63 × 10⁻³⁴ J s; c = 3.0 × 10⁸ m s⁻¹; m_H = 1.67 × 10⁻²⁷ kg.

(a) Energy and momentum of each photon

1. Energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸)/(6.328 × 10⁻⁷) = 3.14 × 10⁻¹⁹ J = 3.14 × 10⁻¹⁹/1.6 × 10⁻¹⁹ = 1.96 eV.
2. Momentum: p = h/λ = (6.63 × 10⁻³⁴)/(6.328 × 10⁻⁷) = 1.05 × 10⁻²⁷ kg m s⁻¹ (= E/c ✓).

(b) Number of photons per second

1. P = N E ⇒ N = P/E = (9.42 × 10⁻³)/(3.14 × 10⁻¹⁹) = 3.0 × 10¹⁶ photons per second.

(c) Speed of a hydrogen atom with the same momentum

1. p = m_H v ⇒ v = p/m_H = (1.05 × 10⁻²⁷)/(1.67 × 10⁻²⁷) = 0.63 m s⁻¹ (about 63 cm s⁻¹).

• Observation: although 3 × 10¹⁶ photons arrive per second, each carries a tiny momentum; a hydrogen atom moving at a leisurely 0.63 m s⁻¹ has as much momentum as a visible-light photon — which is why light exerts only a very small radiation pressure.
• Colour: 632.8 nm is the red line of the He–Ne laser (photon energy 1.96 eV).

Conclusion: (a) each photon has energy 3.14 × 10⁻¹⁹ J (1.96 eV) and momentum 1.05 × 10⁻²⁷ kg m s⁻¹; (b) 3.0 × 10¹⁶ photons arrive per second; (c) a hydrogen atom needs a speed of 0.63 m s⁻¹ to have the same momentum.

Concept and context. Einstein’s photoelectric equation gives K_max = hν − φ₀ and, since K_max = eV₀, V₀ = (h/e) ν − φ₀/e. Hence the graph of the cut-off (stopping) voltage V₀ against the frequency ν of the incident light is a straight line with slope h/e and an intercept on the frequency axis equal to the threshold frequency ν₀ = φ₀/h (NCERT Fig. 11.5). This was the basis of Millikan’s precise measurement of Planck’s constant and the proof of Einstein’s equation. Knowing e, the slope immediately yields h.

Given: slope = ΔV₀/Δν = h/e = 4.12 × 10⁻¹⁵ V s; e = 1.6 × 10⁻¹⁹ C.

1. Relation: V₀ = (h/e)ν − φ₀/e, so slope = h/e.
2. Planck’s constant: h = e × slope = (1.6 × 10⁻¹⁹ C) × (4.12 × 10⁻¹⁵ V s).
3. h = 6.59 × 10⁻³⁴ C V s = 6.59 × 10⁻³⁴ J s (since C V = J).

h = e × (slope) ≈ 6.6 × 10⁻³⁴ J s

• Check: the accepted value is 6.63 × 10⁻³⁴ J s (so the slope of the ideal line is 4.14 × 10⁻¹⁵ V s) — the experimental value agrees to within 1 %.
• Significance: the slope is the same for every metal (universal constant h/e); only the intercept ν₀ differs from metal to metal (it depends on φ₀) — different metals give parallel lines.

Conclusion: Planck’s constant h = e × slope = 6.59 × 10⁻³⁴ J s (≈ 6.6 × 10⁻³⁴ J s).

Concept and context. The threshold frequency ν₀ is the minimum frequency of the incident radiation that can eject electrons from a metal: hν₀ = φ₀ (work function). For ν > ν₀ the emitted electrons have the maximum kinetic energy K_max = h(ν − ν₀) (Einstein’s equation), and the cut-off (stopping) voltage is the retarding potential that just stops them: eV₀ = K_max = h(ν − ν₀) (NCERT Fig. 11.5). Since 8.2 × 10¹⁴ Hz > 3.3 × 10¹⁴ Hz, photoemission occurs.

Given: ν₀ = 3.3 × 10¹⁴ Hz; ν = 8.2 × 10¹⁴ Hz; h = 6.63 × 10⁻³⁴ J s; e = 1.6 × 10⁻¹⁹ C.

1. Einstein’s equation: eV₀ = hν − hν₀ = h(ν − ν₀).
2. ν − ν₀ = (8.2 − 3.3) × 10¹⁴ = 4.9 × 10¹⁴ Hz.
3. V₀ = h(ν − ν₀)/e = (6.63 × 10⁻³⁴ × 4.9 × 10¹⁴)/(1.6 × 10⁻¹⁹) = (3.25 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 2.03 V.

V₀ = h(ν − ν₀)/e ≈ 2.0 V

• Check: (h/e)(ν − ν₀) = 4.14 × 10⁻¹⁵ V s × 4.9 × 10¹⁴ Hz = 2.03 V ✓.
• Increasing the intensity would not change this cut-off voltage.

Conclusion: the cut-off voltage is ≈ 2.0 V (2.03 V).

Concept and context. Photoelectric emission occurs only if the energy of a single incident photon, E = hν = hc/λ, is at least equal to the work function φ₀ of the metal: hν ≥ φ₀ (equivalently ν ≥ ν₀ or λ ≤ λ₀ = hc/φ₀, the threshold wavelength). If hν < φ₀ no electron can escape, however intense the light and however long it shines (NCERT Section 11.6) — an electron absorbs only one photon at a time. A convenient relation: E (eV) = 1240/λ (nm).

1. Photon energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸)/(330 × 10⁻⁹) = 6.03 × 10⁻¹⁹ J = 6.03 × 10⁻¹⁹/1.6 × 10⁻¹⁹ = 3.77 eV (= 1240/330).
2. Compare with the work function: E = 3.77 eV < φ₀ = 4.2 eV.
3. Conclusion: the photon energy is insufficient to liberate an electron: no photoelectric emission takes place.
4. Check by threshold wavelength: λ₀ = hc/φ₀ = 1240/4.2 = 295 nm; the incident wavelength 330 nm > λ₀ (≈ 295 nm), i.e. the frequency 9.1 × 10¹⁴ Hz < ν₀ = 1.01 × 10¹⁵ Hz.

E = 3.77 eV < φ₀ = 4.2 eV ⇒ no photoelectric emission

• What would work? Ultraviolet light of wavelength shorter than 295 nm (e.g. 250 nm: E = 4.96 eV ⇒ K_max = 0.76 eV).
• Why not a stronger beam? Increasing the intensity increases the number of photons but not the energy of each; with the wave picture one would expect emission eventually — but experiments show none (the failure of the wave theory, NCERT Section 11.5).

Conclusion: the photon energy (3.77 eV) is less than the work function (4.2 eV), so the metal will not give photoelectric emission for radiation of wavelength 330 nm.

Concept and context. Einstein’s equation relates the incident frequency, the threshold frequency and the maximum kinetic energy of the photoelectrons: hν = hν₀ + ½ m v²_max, i.e. K_max = ½ m v²_max = h(ν − ν₀). Given the incident frequency and the maximum speed of the emitted electrons we can find ν₀ (and hence the work function φ₀ = hν₀).

Given: ν = 7.21 × 10¹⁴ Hz; v_max = 6.0 × 10⁵ m s⁻¹; m = 9.11 × 10⁻³¹ kg; h = 6.63 × 10⁻³⁴ J s.

1. Maximum kinetic energy: K_max = ½ m v²_max = ½ × 9.11 × 10⁻³¹ × (6.0 × 10⁵)² = ½ × 9.11 × 10⁻³¹ × 3.6 × 10¹¹ = 1.64 × 10⁻¹⁹ J (≈ 1.02 eV).
2. Photon energy: hν = 6.63 × 10⁻³⁴ × 7.21 × 10¹⁴ = 4.78 × 10⁻¹⁹ J.
3. Work function: φ₀ = hν − K_max = 4.78 × 10⁻¹⁹ − 1.64 × 10⁻¹⁹ = 3.14 × 10⁻¹⁹ J (≈ 1.96 eV).
4. Threshold frequency: ν₀ = φ₀/h = (3.14 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 4.74 × 10¹⁴ Hz (≈ 4.7 × 10¹⁴ Hz).

ν₀ = ν − m v²_max/(2h) ≈ 4.7 × 10¹⁴ Hz

• Check: ν = 7.21 × 10¹⁴ Hz > ν₀ = 4.74 × 10¹⁴ Hz ✓ (emission occurs); the metal responds to visible (red) light like the alkali metals.
• The speed 6 × 10⁵ m s⁻¹ is non-relativistic (0.2 % of c).

Conclusion: the threshold frequency for photoemission is ≈ 4.7 × 10¹⁴ Hz (4.74 × 10¹⁴ Hz).

Concept and context. By Einstein’s photoelectric equation the photon energy hν = hc/λ is shared between the work function φ₀ and the maximum kinetic energy of the photoelectrons; the maximum kinetic energy is measured by the stopping potential: K_max = eV₀. Hence φ₀ = hc/λ − eV₀. Working in electron-volts: hc = 1240 eV nm, and eV₀ numerically equals V₀ in volts.

Given: λ = 488 nm; V₀ = 0.38 V; hc = 1240 eV nm (= 1.99 × 10⁻²⁵ J m).

1. Photon energy: E = hc/λ = 1240/488 = 2.54 eV (= 4.07 × 10⁻¹⁹ J).
2. Maximum kinetic energy: K_max = eV₀ = 0.38 eV.
3. Work function: φ₀ = E − K_max = 2.54 − 0.38 = 2.16 eV (≈ 3.46 × 10⁻¹⁹ J).

φ₀ = hc/λ − eV₀ = 2.54 − 0.38 = 2.16 eV

• Interpretation: φ₀ ≈ 2.2 eV is typical of alkali metals such as caesium (2.14 eV) or potassium (2.3 eV) which respond to visible light; the blue-green light of the argon laser has enough energy.
• Check: the stopping potential for a green line of wavelength 546 nm for this metal would be 1240/546 − 2.16 = 0.11 V — decreasing with the wavelength as expected.

Conclusion: the work function of the emitter material is 2.16 eV.

Concept and context. Louis de Broglie (1924) proposed that a moving particle of mass m and speed v has an associated matter wave of wavelength λ = h/p = h/(mv) (NCERT Eq. 11.5), where h = 6.63 × 10⁻³⁴ J s is Planck’s constant. The wavelength is smaller for a heavier or faster particle. For ordinary macroscopic objects, h is so small compared with their momentum that λ is far smaller than any measurable length, which is why no wave-like behaviour is noticed in everyday life (NCERT: the wave character is significant only in the sub-atomic domain).

Given: h = 6.63 × 10⁻³⁴ J s.

1. (a) Bullet: m = 0.040 kg, v = 1.0 km s⁻¹ = 1000 m s⁻¹ ⇒ p = mv = 40 kg m s⁻¹; λ = h/p = 6.63 × 10⁻³⁴/40 = 1.7 × 10⁻³⁵ m.
2. (b) Ball: m = 0.060 kg, v = 1.0 m s⁻¹ ⇒ p = 0.060 kg m s⁻¹; λ = 6.63 × 10⁻³⁴/0.060 = 1.1 × 10⁻³² m.
3. (c) Dust particle: m = 1.0 × 10⁻⁹ kg, v = 2.2 m s⁻¹ ⇒ p = 2.2 × 10⁻⁹ kg m s⁻¹; λ = 6.63 × 10⁻³⁴/(2.2 × 10⁻⁹) = 3.0 × 10⁻²⁵ m.

• Meaning: these wavelengths are 10²⁰–10¹⁰ times smaller than a proton (~10⁻¹⁵ m) or an atom (10⁻¹⁰ m) and are quite beyond measurement; no diffraction or interference of such objects can ever be observed, so classical mechanics is perfectly adequate for them. Compare the electron, whose wavelength 0.135 nm equals X-ray wavelengths and can be diffracted by crystals.
• Rule: λ ∝ 1/(mv): smaller for heavier and faster objects; the bullet (fastest, heaviest) has the smallest wavelength here.

Conclusion: the de Broglie wavelengths are (a) 1.7 × 10⁻³⁵ m, (b) 1.1 × 10⁻³² m, (c) 3.0 × 10⁻²⁵ m — all immeasurably small, so macroscopic objects show no wave-like behaviour.

Concept and context. The de Broglie relation λ = h/p was proposed for material particles; its validity can be tested only by experiment. However, it is interesting that it is also satisfied by a photon, the quantum of electromagnetic radiation: a photon of frequency ν and wavelength λ carries energy E = hν and momentum p = E/c = hν/c (NCERT Section 11.7: Einstein’s result that a light quantum has momentum hν/c). The de Broglie wavelength of the photon, defined as h/p, should therefore equal the wavelength of the electromagnetic wave.

Proof

1. Wave relation: for electromagnetic radiation in vacuum c = νλ ⇒ λ = c/ν.
2. Photon momentum: from special relativity (or Einstein’s photon concept) for a particle of zero rest mass E = pc ⇒ p = E/c = hν/c.
3. de Broglie wavelength: λ_dB = h/p = h/(hν/c) = c/ν.
4. Compare: λ_dB = c/ν = λ. Hence the de Broglie wavelength of a photon equals the wavelength of the electromagnetic radiation of which it is the quantum. Q.E.D.

λ_dB = h/p = h/(hν/c) = c/ν = λ

• Example: for a photon of green light, λ = 500 nm, p = h/λ = 1.33 × 10⁻²⁷ kg m s⁻¹ and E = 2.5 eV: the “particle” attribute (momentum) and the “wave” attribute (λ) are connected by the same relation h = pλ.
• Significance: the dual nature of radiation and matter is described by one relation λ = h/p, with Planck’s constant as the bridge.

Conclusion: with p = hν/c for a photon, the de Broglie wavelength λ = h/p = c/ν, which is exactly the wavelength of the electromagnetic radiation.