Question

When you irradiate a metal with light of wavelength 447 nm in an investigation of the photoelectric effect, you discover that a potential difference of 1.43 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts? energy of a photon: eV Find the work function of the irradiated metal in electron volts. work function:

Answer 1

ENERGY OF PHOTON:-

Photon energy is given by E = hf.

where h is Planck's constant and f is frequency of the photon.

Since we are given the wavelength rather than the frequency, we solve the familiar relationship c = fλ for the frequency, yielding

E = h*c/λ

E = 6.63 * 10^-34 * 3 * 10^8 / (447 * 10^-9)

E = 4.45 * 10^-19 J

Converting to eV, the energy of the photon is

E = 4.45 * 10^-19/1.6 * 10^-19

E = 2.78 eV

WORK FUNCTION:-

now, to find work function we already have the kinetic energy of the ejected electron is 1.43 eV and the energy associated with the photon of lights is 2.78 eV

thus,

work function = Total energy of photon - kinetic energy of ejected electron

work function = 2.78 - 1.43 = 1.35 eV

Final Answers :-

Energy of a photon = 2.78 eV

Work function of irradiated metal = 1.35 eV