Orange light with a wavelength of 6.00x102nm is directed at a metallic surface with a work function of 1.60eV. Calculate:

Question

a) The maximum kinetic energy of the emitted electrons
b) Their maximum speed
c) The cutoff potential necessary to stop these electrons (V=Ek/e)

Answer 1

wait...did u mean a wavelength of "6.00 x 10^2"?

if so...these are my answers

a) first convert the numbers

6.00 x 10^2 nm = 6 x 10^-7 m
1.60 eV = 2.56 x 10^-19

then, plug in the numbers in this equation:

KE = hc/l - w

KE = (6.63E-34)(3E8)/(6E-7) - (2.56E-19) = 7.55E-20 J

b) for the velocity, use the ususal kinetic energy equation and solve for velocity

KE = (1/2)mv^2

then...

v = squ-root(2 x KE/m)

(assuming the mass is an electron...)

v = squ-root(2 x (7.55E-20)/(9.1094E-31)) = 4.0714E5 m/s

c) i have no idea how to do...srry bout that mate...