The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
For photoemission energy ‘E’ of incident radiation must be more than or equal to work function ‘ϕ0’ of the metal.
E 
ϕ0 (The Condition for photoelectric emission)
E = hv = hc/λ
Where h = plank’s constant = 6.63 × 10-34 J. sec
ν = frequency of radiation
c = speed of light = 3 × 108 m/s
λ = wavelength of radiation
E = 6.63 × 10-34 × 3 × 108 / 330 × 10-9
= 6.018 × 10-19 J = 6.018 × 10-19/1.6 × 10-19 eV
= 3.76 eV
ϕ0 = 4.2 eV as per question.
E < ϕ0, therefore photoemission will not take place.
No photoemission will take place as work function is more than the photon energy of the incident radiation.
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