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1)

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


A) <2.8×109m

B) 2.8×109m

C) 2.8×1012m

D) <2.8×1010m

Answer:

Option B

Explanation:

Given : work function Φ of metal = 2.28 eV

Wavelength of light λ = 500 nm = 500 x 10-9m

KEmax=hcλϕ

KEmax=6.6×1034×3×1085×1072.82

KEmax=2.482.28=0.2eV

λmin=hp=h2m(KE)max

203×10342×9×1031×0.2×1.6×1019

λmin=259×109 = 2.80 x 10-9nm

.'. λ ≥ 2.8 x 10-9m