Dual Nature of Radiation and Matter

1 exercises12 questions solved

Exercise 11.1Photoelectric Effect and de Broglie Wavelength

Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.

Solution

V = 30 kV = 30 × 10³ V Maximum energy of X-rays = kinetic energy of electrons: hf_max = eV (a) f_max = eV/h = (1.6 × 10⁻¹⁹ × 30 × 10³)/(6.626 × 10⁻³⁴) = (4.8 × 10⁻¹⁵)/(6.626 × 10⁻³⁴) = 7.24 × 10¹⁸ Hz (b) λ_min = c/f_max = 3 × 10⁸ / (7.24 × 10¹⁸) = 4.14 × 10⁻¹¹ m ≈ 0.0414 nm

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 is the (a) maximum kinetic energy of the emitted electrons, (b) stopping potential, and (c) maximum speed of the emitted photoelectrons?

Solution

φ = 2.14 eV = 2.14 × 1.6 × 10⁻¹⁹ J, f = 6 × 10¹⁴ Hz hf = 6.626 × 10⁻³⁴ × 6 × 10¹⁴ = 3.976 × 10⁻¹⁹ J = 2.485 eV (a) KE_max = hf – φ = 2.485 – 2.14 = 0.345 eV = 0.345 × 1.6 × 10⁻¹⁹ = 5.52 × 10⁻²⁰ J (b) Stopping potential: eV₀ = KE_max = 0.345 eV V₀ = 0.345 V (c) Maximum speed: ½mv² = KE_max v = √(2 × KE_max/m) = √(2 × 5.52 × 10⁻²⁰/9.11 × 10⁻³¹) = √(1.212 × 10¹¹) = 3.48 × 10⁵ m/s

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

Solution

Cut-off (stopping) voltage V₀ = 1.5 V Maximum kinetic energy: KE_max = eV₀ = 1.6 × 10⁻¹⁹ × 1.5 = 2.4 × 10⁻¹⁹ J = 1.5 eV

Monochromatic 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?

Solution

λ = 632.8 nm = 632.8 × 10⁻⁹ m, P = 9.42 mW = 9.42 × 10⁻³ W (a) Energy of each photon: E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(632.8 × 10⁻⁹) = 1.988 × 10⁻²⁵/632.8 × 10⁻⁹ = 3.14 × 10⁻¹⁹ J = 1.96 eV Momentum: p = h/λ = 6.626 × 10⁻³⁴/632.8 × 10⁻⁹ = 1.047 × 10⁻²⁷ kg·m/s (b) Number of photons per second: N = P/E = 9.42 × 10⁻³/3.14 × 10⁻¹⁹ = 3 × 10¹⁶ photons/s (c) Speed of hydrogen atom with same momentum: m_H = 1.67 × 10⁻²⁷ kg v = p/m_H = 1.047 × 10⁻²⁷/1.67 × 10⁻²⁷ = 0.627 m/s

The energy flux of sunlight reaching the surface of the earth is 1.388 × 10³ W/m². 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.

Solution

Energy flux = 1.388 × 10³ W/m², λ = 550 nm = 550 × 10⁻⁹ m Energy per photon: E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(550 × 10⁻⁹) = 1.988 × 10⁻²⁵/550 × 10⁻⁹ = 3.61 × 10⁻¹⁹ J Number of photons per m² per second: N = Energy flux/E = 1.388 × 10³/3.61 × 10⁻¹⁹ = 3.84 × 10²¹ photons/m²/s

In 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.

Solution

Einstein's photoelectric equation: eV₀ = hf – φ V₀ = (h/e)f – φ/e Slope of V₀ vs f graph: h/e = 4.12 × 10⁻¹⁵ V·s h = e × 4.12 × 10⁻¹⁵ = 1.6 × 10⁻¹⁹ × 4.12 × 10⁻¹⁵ = 6.59 × 10⁻³⁴ J·s This matches the known value of Planck's constant (6.626 × 10⁻³⁴ J·s) ✓

A 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. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere? λ = 589 nm.

Solution

P = 100 W, λ = 589 nm = 589 × 10⁻⁹ m (a) Energy per photon: E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(589 × 10⁻⁹) = 3.37 × 10⁻¹⁹ J = 2.11 eV (b) Rate of photons: N/t = P/E = 100/(3.37 × 10⁻¹⁹) = 2.97 × 10²⁰ ≈ 3 × 10²⁰ photons/s

The 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.

Solution

f₀ = 3.3 × 10¹⁴ Hz, f = 8.2 × 10¹⁴ Hz Using Einstein's photoelectric equation: eV₀ = h(f – f₀) V₀ = h(f – f₀)/e = [6.626 × 10⁻³⁴ × (8.2 – 3.3) × 10¹⁴] / (1.6 × 10⁻¹⁹) = [6.626 × 10⁻³⁴ × 4.9 × 10¹⁴] / (1.6 × 10⁻¹⁹) = [32.47 × 10⁻²⁰] / (1.6 × 10⁻¹⁹) = 2.03 V ≈ 2.0 V

Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Solution

f = 7.21 × 10¹⁴ Hz, v_max = 6.0 × 10⁵ m/s Using: hf = φ + ½mv_max² φ = hf – ½mv_max² = 6.626 × 10⁻³⁴ × 7.21 × 10¹⁴ – ½ × 9.11 × 10⁻³¹ × (6.0 × 10⁵)² = 4.777 × 10⁻¹⁹ – ½ × 9.11 × 10⁻³¹ × 3.6 × 10¹¹ = 4.777 × 10⁻¹⁹ – 1.64 × 10⁻¹⁹ = 3.137 × 10⁻¹⁹ J Threshold frequency: f₀ = φ/h = 3.137 × 10⁻¹⁹/6.626 × 10⁻³⁴ = 4.73 × 10¹⁴ Hz

Q10

What is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the 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?

Solution

de Broglie wavelength: λ = h/mv = h/p h = 6.626 × 10⁻³⁴ J·s (a) Bullet: m = 0.040 kg, v = 1000 m/s λ = 6.626 × 10⁻³⁴/(0.040 × 1000) = 6.626 × 10⁻³⁴/40 = 1.66 × 10⁻³⁵ m (Extremely small — wave nature completely negligible) (b) Ball: m = 0.060 kg, v = 1.0 m/s λ = 6.626 × 10⁻³⁴/(0.060 × 1.0) = 1.10 × 10⁻³² m (c) Dust particle: m = 1.0 × 10⁻⁹ kg, v = 2.2 m/s λ = 6.626 × 10⁻³⁴/(1.0 × 10⁻⁹ × 2.2) = 6.626 × 10⁻³⁴/2.2 × 10⁻⁹ = 3.01 × 10⁻²⁵ m

Q11

An electron, an alpha particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?

Solution

KE = p²/(2m) → p = √(2mKE) de Broglie wavelength: λ = h/p = h/√(2mKE) For same KE: λ ∝ 1/√m Larger mass → smaller λ Masses: • Electron: me = 9.11 × 10⁻³¹ kg (lightest) • Proton: mp = 1.67 × 10⁻²⁷ kg • Alpha particle: mα = 4 × mp = 6.68 × 10⁻²⁷ kg (heaviest) Alpha particle has the largest mass → smallest λ The alpha particle has the shortest de Broglie wavelength.

Q12

What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the r.m.s. speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u, 1 u = 1.66 × 10⁻²⁷ kg, Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹)

Solution

N₂ molecule: M = 28.0152 u = 28 × 1.66 × 10⁻²⁷ = 4.648 × 10⁻²⁶ kg T = 300 K, k_B = 1.38 × 10⁻²³ J/K rms speed: v_rms = √(3k_BT/m) = √(3 × 1.38 × 10⁻²³ × 300 / 4.648 × 10⁻²⁶) = √(1.242 × 10⁻²⁰ / 4.648 × 10⁻²⁶) = √(2.672 × 10⁵) = 516.9 m/s de Broglie wavelength: λ = h/(mv_rms) = 6.626 × 10⁻³⁴/(4.648 × 10⁻²⁶ × 516.9) = 6.626 × 10⁻³⁴/(2.402 × 10⁻²³) = 2.76 × 10⁻¹¹ m = 0.028 nm

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