Electrons are ejected from a metallic surface with speeds of up to 4.60x105 m/s when light with a wave length of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface? (

The solution to the question above is explained below:

Explanation:

(a) The work function of the surface is:

The work function of a metal, Φ, it's the minimum amount of energy required to remove electron from the conduction band and remove it to outside the metal. It is typically exhibited in units of eV (electron volts) or J (Joules). Work Function, is the minimum thermodynamic work.

hf = hc /λ = φ + Kmax = φ + 1 /2 mev ²max

where,   φ=work function of a metal

               eV=electron volts

              J= Joules

             λ=the Plank constant 6.63 x 10∧-34 J s

              f = the frequency of the incident light in hertz

              ∧= signifies raised to the power in the solution

Kmax= the maximum kinetic energy of the emitted electrons in joules

φ= ( hc /λ -1/2mev ²max= 6.63 × 10∧-34Js × 3.00 × 10∧8 m/s ÷ 625 × 10∧-9) -

 (1/2 × 9.11 × 10∧-31 kg × 4.60 × 10∧5 m/s) = 2.21 × 10∧-19 J

ans= 1.38 eV

(b) The cutoff frequency for this surface is:

The cutoff frequency is the minimum frequency that is required for the emission of electrons from a metallic surface at which energy flowing through the metallic surface begins to be reduced  rather than passing through.

Light at the cutoff frequency only barely supplies enough energy to overcome the work function. The value of the cutoff frequency is in unit hertz (Hz)

hfcut = φ

fcut = φ /h

ans= 334 THz

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )