Question:
light of wavelength 500nm is incident on a metal with work function 2.28ev the se Broglie wavelength of emittied electron is what
Answer:
E = W0 + eV
So, hc/λ = W0 + eV
(6.6 x 10-34 x 3 x 108)/(500 x 10-9 x 1.6 x 10-19) = 2.28 + eV
So, V <= 2
de-Broglie wavelength of electron
λ = 12.27/√V
So, λ >= 2.8 x 10-9 metres.
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