Energy required to remove an electron from an aluminum surface is 4.2 eV, lf light of wavelength 2000 A falls on the surface, the velocity of fastest electron erected from the surface is
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a
2.8 x1018 m/s
b
2.5 x1013 m/s
c
6.7 x1018 m/s
d
none of these
answer is D.
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Detailed Solution
12mv2=hcλ−W0 12mv2=6⋅6×10−343×10∘2000×10−10×1⋅6×10−19−4⋅2eV =6⋅2eV−4⋅2eV=2eV =2×1⋅6×10−19=3⋅2×10−19J Now v=2×3⋅2×10−199⋅1×10−31=0⋅703×106m/s