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Question

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

A
2.8×1012m
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B
<2.8×1010m
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C
<2.8×109m
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D
2.8×109m
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Solution

The correct option is D 2.8×109m
From photoelectric effect, hν=W0+eV

or hcλ=W0+eV

or (6.6×1034)×(3×108)500×109
=2.28×1.6×1019+eV
or eV=0.31×1019 or V=0.2

Here V should be V0.2

The de Broglie wavelength of electron , λ=12.27×1010V

So, λ2.8×109m

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