Question

Please help.

21 A piece of metal has a cutoff wavelength of cutofi 450 nm. Consider illuminating this piece of metal with a different waverlenght of light: a 2400 nm beam. What is the maximum kinetic energy of ejected electrons?

Answer 1

Q21. According to photoelectric effect equation,

K.E._max = (h * c) * (1/$\lambda$ - 1/$\lambda$_cutoff)

where

K.E._max = maximum kinetic energy of an electron

h = Planck's constant = 6.626 x 10^-34 J.s

c = speed of light = 3.0 x 10⁸ m/s

$\lambda$ = wavelength of light = 400 nm = 400 x 10^-9 m

$\lambda$_cutoff = cutoff wavelength of metal = 450 nm = 450 x 10^-9 m

substituting the values,

K.E._max = [(6.626 x 10^-34 J.s) * (3.0 x 10⁸ m/s)] * (1/400 x 10^-9 m - 1/450 x 10^-9 m)

K.E._max = [(6.626 x 10^-34 J.s) * (3.0 x 10⁸ m/s)] * (2.78 x 10⁵ m^-1)

K.E._max = 5.5 x 10^-20 J

This is the maximum kinetic energy of an ejected electron.

K.E._max per mole electrons = (K.E._max for one electron) * (Avogadro's number)

K.E._max per mole electrons = (5.5 x 10^-20 J/electron) * (6.022 x 10²³ electrons/mol)

K.E._max per mole electrons = 33251.5 J/mol

K.E._max per mole electrons = 33.3 kJ/mol