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Question:

Light of wavelength 500 nm is incident on a metal with a work function 2.28 eV. The de Broglie wavelength of the emitted electron is:

≤2.8×10⁻¹⁴m

<2.8×10⁻¹²m

≥2.8×10⁻⁹m

<2.8×10⁻⁹m

Solution:

Energy of the photon: E = 12400/5000 = 2.48 eV
Work function: φ = 2.28 eV
KEmax = E − φ = 2.48 − 2.28 = 0.2 eV
For electron, λmin = h/√2m(KE)max = 28 Å
Hence, λ > 28 Å = 2.8 × 10⁻⁹ m