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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

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< 2.8 × 10^-9 m

≤ 2.8 × 10^-12 m

< 2.8 × 10^-10 m

≥ 2.8 × 10^-9 m

answer is D.

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Detailed Solution

$K \leq \frac{1240}{500} eV - 2.28 eV = 0.2 eV$

$λ \geq \frac{12.27 \overset{0}{A}}{\sqrt{0.2}} \approx 2.8 nm$

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