A singly charged positive particle initially has kinetic energy 3.0 eV and de Broglie wavelength 0.50 nm. It then passes through an additional accelerating potential difference of 12 V, oriented so that its kinetic energy increases. If its final de Broglie wavelength is approximately N pm, find N to the nearest integer.
Physics · Photoelectric effect and dual nature of matter · debroglie duel nature of light
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Question

A singly charged positive particle initially has kinetic energy 3.0 eV and de Broglie wavelength 0.50 nm. It then passes through an additional accelerating potential difference of 12 V, oriented so that its kinetic energy increases. If its final de Broglie wavelength is approximately N pm, find N to the nearest integer.

Physics · Photoelectric effect and dual nature of matter · debroglie duel nature of light

G1 Physics · Accepted Answer

Because the particle is singly charged, acceleration through 12 V adds 12 eV of kinetic energy. Initial kinetic energy: K₁ = 3.0 eV. Final kinetic energy: K₂ = 3.0+12.0 = 15.0 eV. For fixed mass, λ is inversely proportional to √K. Hence, λ₂ = λ₁√(K₁ ÷ K₂) = 0.50√(3 ÷ 15) nm = 0.50 ÷ √5 nm ≈ 0.2236 nm. Since 1 nm = 1000 pm, λ₂ ≈ 223.6 pm. To the nearest integer, N = 224.