TP 30.28

Given: W_0,Pt = 6.35 eV = 10.16 x 10^-19 J

W_0,Fe = 4.50 eV = 7.20 x 10^-19 J

f = 1.88 x 10¹⁵ Hz

Find: (a) K_max - ?

The less energy is spent to free electron from the metal, the greater maximum kinetic energy the ejected electrons have.

Thus electrons ejected from iron will have higher maximum kinetic energy.

(b) K_max - ?

K_max = h f - W₀

K_{max, Pt} = (6.63 x 10^-34 J-s)( 1.88 x 10¹⁵ Hz) – (10.16 x 10^-19 J) = 2.30 x 10^-19 J

K_{max, Fe} = (6.63 x 10^-34 J-s)( 1.88 x 10¹⁵ Hz) – (7.20 x 10^-19 J) = 5.26 x 10^-19 J