A light wave has electric field E = 10³ cos ((2π x)/(5×10⁻⁷) - 2π×6×10¹⁴ t)x̂ N/C. When this radiation is incident on a metal surface whose work function is 2 eV, what stopping potential is obtained for the emitted photoelectrons? Use E (in eV) = (12375)/(λ (in Å)).

Question

Accepted Answer

0.48 V. The energy of the incident photons can be calculated using the wavelength derived from the electric field equation, which gives a wavelength of 5000 Å. Using the formula for photon energy, we find it to be approximately 2.48 eV. Subtracting the work function of 2 eV from this energy yields a stopping potential of 0.48 V for the emitted photoelectrons.

A light wave has electric field E = 10³ cos ((2π x)/(5×10⁻⁷) - 2π×6×10¹⁴ t)x̂ N/C. When this radiation is incident on a metal surface whose work function is 2 eV, what stopping potential is obtained for the emitted photoelectrons? Use E (in eV) = (12375)/(λ (in Å)).

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