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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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a
≥2.8×10-9 m
b
≤2.8×10-12 m
c
<2.8×10-10 m
d
<2.8×10-9 m

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detailed solution

Correct option is A

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted electron isKmax=hcλ-ϕ0where λ is the wavelength of incident light and ϕ0 is the work function. Here, λ=500nm,hc=1240eVnm and ϕ0=2.28eV∴  Kmax=1240eVnm500nm-2.28eV=2.48eV-2.28eV=0.2eVThe de Broglie wavelength of the emitted electron is λmin=h2mKmaxwhere h is the Planck's constant and m is the mass of the electron. As h=6.6×10-34 J s,  m=9×10-31 kg and Kmax=0.2eV=0.2×1.6×10-19 J∴  λmin=6.6×10-34 J s29×10-31 kg0.2×1.6×10-19 J=6.62.4×10-9 m=2.8×10-9 m So, λ≥2.8×10-9 m

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