Example 1 · level 1
Find the energy of a photon of wavelength 620 nm, in eV (correct to 2 decimal places). Use hc = 1240 eV nm.
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- E = 1240 ÷ 620 = 2 eV.
Answer: 2 eV
CLASS 11–12 PHYSICS · DUAL NATURE OF RADIATION AND MATTER
Photons, photoelectric effect and de Broglie wavelength
Photon energy E = hc/λ ≈ 1240/λ(nm) eV. Photoelectric: K_max = E − φ. Electron accelerated through V volts: λ ≈ 1.227/√V nm = 1227/√V pm.
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Worked examples
Example 2 · level 2
Light of wavelength 310 nm falls on sodium (work function 2.75 eV). Find the maximum kinetic energy of the emitted photoelectrons, in eV (correct to 2 decimal places). Use hc = 1240 eV nm.
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- E = 4 eV.
- K_max = 4 − 2.75 = 1.25 eV.
- Stopping potential = 1.25 V.
Answer: 1.25 eV
Example 3 · level 3
An electron is accelerated from rest through 2,500 V. Find its de Broglie wavelength in picometres (pm), correct to 2 decimal places. Use λ = 1.227/√V nm.
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- λ = 1227 ÷ 50 = 24.54 pm (= 0.0245 nm).
Answer: 24.54 pm