Question

Photons with a frequency of 5.6*10^14 s-1 are required to eject electrons from potassium metal. what is the kinetic energy of the ejected electrons when 450 nm photons shine on the metal?

Answer 1

step 1: find the energy of photon that is shine

we have:

wavelength = 4.5*10^-7 m

we have below equation to be used:

Energy = Planck constant*speed of light/wavelength

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(4.5*10^-7 m)

= 4.417*10^-19 J

E = 4.417*10^-19 J

step 2: find the threshold energy

we have:

frequency = 5.6*10^14 s-1

we have below equation to be used:

Energy = Planck constant*frequency

=(6.626*10^-34 J.s)*(5.6*10^14) s-1

= 3.711*10^-19 J

Eo = 3.711*10^-19 J

step 3:

KE = E - Eo

= 4.417*10^-19 J - 3.711*10^-19 J

= 7.06*10^-20 J

Answer: 7.06*10^-20 J