The difference between the incident energy and threshold energy for an electron in a photoelectric effect experiment is 5 eV. The de Broglie wavelength of the electron is
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a
6.6 × 10−91456 m
b
6.6 × 10−9145.6 m
c
6.6 × 10−91664 m
d
6.6 × 10−9166.4 m
answer is B.
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Detailed Solution
KE of e− ejected = Energy of incident quantum - threshold energy = 5 eV==5×1.6×10-19λ=hmV = h2mK.E=6.6 × 10−342 × 9.1 × 10−31 × 5 × 1.6 × 10−19=6.6 × 10−9145.6 m