The energy required to remove an electron from the aluminium surface is φ0=4.2eV. The wavelength of the light is given as λ=2000Å=2×10−7m. Using Einstein's photoelectric equation:
1
2
mvmax2=
hc
λ
−φ0 Given the constants: Planck's constant h=6.6×10−34Js Speed of light c=3×108m∕s Conversion factor 1eV=1.6×10−19J Mass of electron m=9.1×10−31kg The calculation becomes:
1
2
mvmax2=
6.6×10−34×3×108
2×10−7×1.6×10−19
eV Further simplifying:
1
2
mvmax2=
6.2eV−4.2eV
9.1×10−31
vmax2=
2×2×1.6×10−19J
9.1×10−31
vmax2=
6.4
91
×1012⇒vmax=√
6.4
91
×1012 Thus, the velocity of the fastest ejected electron is: vmax=0.838×106⇒vmax=8.4×105m∕s
