For intensity i of a light of wavelength 5000a

Asked by admin @ 17/08/2021 in Physics viewed by 385 People

0. For intensity I of a light of wavelength 5000 À the photoelectron saturation current is 0.40 UA

and stopping potential is 1.36 V. Find the work function of metal.hcHint : E = wo + Kmax,

Answered by admin @ 17/08/2021

Given info : For intensity I of a light of wavelength 5000 Å, the photoelectron saturation current is 0.4μA and stopping potential is 1.36 V.

To find : The work function of metal is ...

solution : incident energy, E = hc/λ

= 12400/λ (in A°) eV

here λ = 5000 A°

so, E = 12400/5000 = 2.48 eV

stopping potential, V = 1.36 volts

so stopping potential energy = making kinetic energy

⇒1.36 eV = Maximum kinetic energy

now using photoelectric effect equation,

E = W₀ + Kmax

⇒2.48 eV = W₀ + 1.36 eV

⇒1.12 eV = W₀

Therefore work function of metal is 1.12 eV.

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For intensity i of a light of wavelength 5000a

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