Physics · Dual Nature of Matter and Radiation

The energy required to remove an electron from an aluminium surface is 4.2 eV. If light of wavelength 2000 Å falls on the surface, the velocity of the fastest ejected photoelectron will be approximately (h = 6.63 × 10 -34 J s, m e = 9.1 × 10 -31 kg):

Verified Solution4 OptionsUpdated: October 2026👁️1 View

Correct answer

8.4 × 10 5 m/s

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The energy required to remove an electron from an aluminium surface is 4.2 eV. If light of wavelength 2000 &Aring; falls on the surface, the velocity of the fastest ejected photoelectron will be approximately (h = 6.63 &times; 10 -34 J s, m e = 9.1 &times; 10 -31 kg):