If the work function of a metal is 4 eV is 5000 A ∘ wavelength of light is incident on the metal. Is there any photoelectric effect?

The emission will happen only if the energy of a photon is more than or equal to the energy possessed by the electron of the metal (work function). The maximum energy state of the emitted electrons depends on the frequency of the incident light but doesn't depend upon the amount of light incident. Work function ϕ o = h c λ o . Given, ϕ o = 4 eV . Threshold wavelength λ o = h c ϕ o = 6 . 62 × 10 - 34 × 3 × 10 8 4 × 1 . 6 × 10 - 19 ⇒ λ o = 3103 × 10 - 10 m . Hence, the maximum wavelength of the light for which the metal will emit electrons is 3103 × 10 - 10 m , so for the given wavelength 5000 × 10 - 10 m no photoelectric effect.

The wavelength of photons in two cases are 4000 A ∘ and 3600 A ∘ , respectively, what is difference in stopping potential for these two?

According to Einstein's equation of photoelectric effect: h c λ = ϕ o + e V o . For case 1 : λ = 4 × 10 - 7 m h c 4 × 10 - 7 = ϕ o + e V o 1 . . . 1 For case 2 : λ = 3 . 6 × 10 - 7 m h c 3 . 6 × 10 - 7 = ϕ o + e V o 2 . . . 2 From the equation 1 and 2 , we get, e V o 2 - V o 1 = h c 3 . 6 × 10 - 7 - h c 4 × 10 - 7 ⇒ V o 2 - V o 1 = 6 . 62 × 10 - 34 × 3 × 10 8 1 . 6 × 10 - 19 1 3 . 6 × 10 - 7 - 1 4 × 10 - 7 ⇒ V o 2 - V o 1 = 0 . 34 volt .