Chapter 11: Dual Nature of Radiation and Matter
Key Formulas:
• Photoelectric: E = hν = hc/λ
• Kmax = hν − W₀
• de Broglie: λ = h/(mv) = h/p
• Stopping potential: eV₀ = Kmax
Q1. The work function of a metal is 2 eV. Find the threshold wavelength and the maximum kinetic energy of photoelectrons when light of wavelength 4000 Å falls on it.
Solution
W₀ = hc/λ₀
λ₀ = hc/W₀ = (6.63 × 10−34)(3 × 108)/(2 × 1.6 × 10−19)
= 19.89 × 10−26/(3.2 × 10−19)
= 6.22 × 10−7 m = 6220 Å
Energy of incident photon: E = hc/λ = (6.63 × 10−34)(3 × 108)/(4 × 10−7) = 4.97 × 10−19 J = 3.11 eV
Kmax = E − W₀ = 3.11 − 2 = 1.11 eV
Q2. Find the de Broglie wavelength of an electron accelerated through a potential difference of 100V.
Solution
λ = 12.27/√V Å (for electrons)
= 12.27/√100 Å
= 12.27/10 Å
= 1.227 Å
Q3. Explain why the photoelectric effect cannot be explained by wave theory of light.
Solution
Wave theory predicts that:
1. Light of any frequency should cause photoelectric effect - but experiments show a threshold frequency exists.
2. Increasing intensity should increase kinetic energy - but KE depends on frequency, not intensity.
3. There should be a time lag - but emission is instantaneous.
These observations contradict wave theory and are explained by Einstein's photon theory where light consists of particles (photons) of energy E = hν.
Q4. The stopping potential for photoelectrons from a metal surface is 3V when light of wavelength 2500 Å is used. Find (i) the work function (ii) threshold wavelength.
Solution
(i) Kmax = eV₀ = 3 eV
E = hc/λ = (6.63 × 10−34)(3 × 108)/(2.5 × 10−7) = 4.97 eV
W₀ = E − Kmax = 4.97 − 3 = 1.97 eV
(ii) λ₀ = hc/W₀ = 12400/1.97 = 6294 Å
Q5. An alpha particle and a proton have the same kinetic energy. Find the ratio of their de Broglie wavelengths.
Solution
λ = h/p = h/√(2mK)
λα/λp = √(mp/mα) = √(1/4) = 1/2
Q6. Light of wavelength 2000 Å falls on a metal of work function 4.2 eV. Find the kinetic energy of the emitted photoelectrons and the stopping potential.
Solution
Energy of photon: E = hc/λ
= (6.63 × 10−34)(3 × 108)/(2 × 10−7)
= 9.945 × 10−19 J = 6.22 eV
Kmax = E − W₀ = 6.22 − 4.2 = 2.02 eV
Stopping potential: V₀ = Kmax/e = 2.02 V
Q7. The de Broglie wavelength associated with a proton is 1.0 × 10−13 m. Find (i) its momentum (ii) its kinetic energy (iii) the accelerating potential. (mp = 1.67 × 10−27 kg)
Solution
(i) p = h/λ = (6.63 × 10−34)/(10−13)
= 6.63 × 10−21 kg·m/s
(ii) K = p²/(2m) = (6.63 × 10−21)²/(2 × 1.67 × 10−27)
= 4.396 × 10−41/(3.34 × 10−27)
= 1.32 × 10−14 J = 82.4 keV
(iii) V = K/q = (1.32 × 10−14)/(1.6 × 10−19) = 82,500 V = 82.5 kV
Q8. Two metals A and B have work functions of 2 eV and 4 eV respectively. Which metal will show photoelectric emission for light of wavelength 4000 Å? Find the maximum kinetic energy for that metal.
Solution
Energy of photon: E = hc/λ
= 12400/4000 = 3.1 eV
Metal A (W₀ = 2 eV): E = 3.1 > 2 → emission occurs
Metal B (W₀ = 4 eV): E = 3.1 < 4 → no emission
Kmax for metal A = 3.1 − 2 = 1.1 eV
Q9. An electron and a proton have the same de Broglie wavelength. Find the ratio of their kinetic energies.
Solution
λ = h/p ⇒ p = h/λ (same for both)
K = p²/(2m)
Ke/Kp = mp/me
= (1.67 × 10−27)/(9.1 × 10−31)
= 1835
Q10. The work function of a metal is 3.3 eV. What is the maximum frequency of electromagnetic radiation that can cause photoelectric emission? What is the corresponding wavelength?
Solution
hν₀ = W₀
ν₀ = W₀/h = (3.3 × 1.6 × 10−19)/(6.63 × 10−34)
= 5.28 × 10−19/(6.63 × 10−34)
= 7.96 × 1014 Hz
λ₀ = c/ν₀ = 12400/3.3 = 3758 Å
Q11. In an experiment, the stopping potential for two different wavelengths λ₁ = 4000 Å and λ₂ = 6000 Å are found to be V₁ and V₂ respectively. Find the ratio V₁/V₂.
Solution
eV₀ = hc/λ − W₀
V₁ = (hc/λ₁ − W₀)/e
V₂ = (hc/λ₂ − W₀)/e
V₁/V₂ = (hc/λ₁ − W₀)/(hc/λ₂ − W₀)
Using hc = 12400 eV·Å:
V₁/V₂ = (12400/4000 − W₀)/(12400/6000 − W₀)
= (3.1 − W₀)/(2.067 − W₀)
The exact ratio depends on W₀. The ratio is not a fixed number without knowing W₀.
Q12. Show graphically how the stopping potential varies with the frequency of incident radiation. What does the slope of the graph represent?
Solution
From Einstein's equation: eV₀ = hν − W₀
V₀ = (h/e)ν − W₀/e
This is a straight line (V₀ vs ν):
• Slope = h/e (Planck's constant / charge of electron)
• x-intercept = ν₀ (threshold frequency)
• y-intercept = −W₀/e (negative)
Slope = h/e = 4.14 × 10−15 V·s
Q13. An α-particle, a proton, and an electron have the same kinetic energy. Find the ratio of their de Broglie wavelengths.
Solution
λ = h/√(2mK)
λα : λp : λe = 1/√(mα) : 1/√(mp) : 1/√(me)
mα = 4mp, mp = 1836me
= 1/√(4mp) : 1/√(mp) : 1/√(me)
= 1/(2√mp) : 1/√mp : 1/√me
= 1/2 : 1 : √(1836)
= 1 : 2 : 85.6 (approximate ratio)