Question

When a metallic surface is illuminated with light of wavelength?

Answer 1

You need to find the work function in order to solve this problem.

Also Planck's constant is also 4.14 * 10^-15 eV.

Max KE = 1.10 eV

lambda = 400 nm = 400 * 10^-9 m

First, find the frequency.

f = c / lambda

f = (3 * 10^8) / (400 * 10^-9 m)

f = 7.5 * 10^14

Max KE = hf - work function

1.10 eV = (4.14 * 10^-15 eV) (7.5 * 10^14) - work function

work function = 2.01 eV

Now find the frequency for the second wavelength.

f = c / lambda

f = (3 * 10^8) / (320 * 10^-9)

f = 9.38 * 10^14

Now plug this into your equation with this frequency and work function you found in the previous problem.

KEmax = hf - work function

KEmax = (4.14 * 10^-15 eV)(9.38 * 10^14) - 2.01 eV

KEmax = 1.87 eV

Hope this helps.