Light of wavelength 250 nm is incident on the surface of cesium. The work function of cesium is 2.1 eV.?

Photoelectric equation:

hf = Φ + KE

Φ = work function (eV)

h = Planck’s constant = 4.1357 × 10^-15 eV • s

KE = maximum kinetic energy of emitted electrons (eV)

c = speed of light = 2.998 x 10^8 m/s

f = minimum frequency

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λ = 250 nm = 2.50 x 10^-7 m

E(photon) = (hc/λ)

= { [(4.1357 × 10^-15 eV • s) * (2.998 x 10^8 m/s)] / (2.50 x 10^-7 m) }

= 4.96 eV

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The photoelectric equation uses hf, which is the same as (hc/λ), so, rearranging the equation to solve for KE,

KE = (hc/λ) - Φ

= (4.96 eV) - (2.1 eV)

= 2.86 eV ………… to 3 significant figures

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I'm actually not sure what significant figures to use. 250 nm has 3 s.f., and 2.1 eV has 2 s.f. Perhaps the final answer should have 2 s.f.