KEmax=λhc−Ψ
⇒KEmax=5001240−2.28
⇒KEmax=2.48−2.28=0.2eV
λmin=2m(KE)maxh=2×9×10−31×0.2×1.6×10−19320×10−34
λmin=925×10−9=2.80×10−9m
So, λ≥2.8×10−9m.
Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is:
NEET UG 2015 — Physics Modern Physics
2015mcqmedium
Light of wavelength 500nm is incident on a metal with work function 2.28eV. The de Broglie wavelength of the emitted electron is:
Official previous-year question
Held on 30 Apr 2015 · Verified 9 Jul 2026.
Options
- A
<2.8×10−9m
- B
≥2.8×10−9m
- C
≤2.8×10−12m
- D
<2.8×10−10m
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