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