Photons of energy 6 eV cause electrons to be emitted from acertain metal with a

Question

Photons of energy 6 eV cause electrons to be emitted from acertain metal with a maximum kinetic energy of 2 eV. If photons oftwice the wavelength are incident on this metal which one of thefollowing statements is true?

a. Electrons will be emitted with amaximum kinetic energy of 8 eV.
b. Electrons will be emitted with amaximum kinetic energy of 10 eV.
c. Electrons will be emitted with amaximum kinetic energy of 20 eV.
d. Electrons will be emitted with amaximum kinetic energy of 1 eV.
e. No electrons will be emitted. Photons of energy 6 eV cause electrons to be emitted from acertain metal with a maximum kinetic energy of 2 eV. If photons oftwice the wavelength are incident on this metal which one of thefollowing statements is true? Electrons will be emitted with amaximum kinetic energy of 8 eV. Electrons will be emitted with amaximum kinetic energy of 10 eV. Electrons will be emitted with amaximum kinetic energy of 20 eV. Electrons will be emitted with amaximum kinetic energy of 1 eV. No electrons will be emitted.

Explanation / Answer

Accroding to Einstein,the maximum kinetic energy for theseliberated electrons is Kmax = hf - -----------(1) where is called the work function of the metal.The workfunction represents the minimum energy with which an electron isbound in the metal and is on the order of a few electronvolts. the energy of the incident photon is hf = 6 eV the maximum kinetic energy of the electrons emitted from themetal surface is Kmax = 2 eV from equation (1) the work function of the metal is = hf - Kmax = 6 - 2 = 4 eV when photons of twice the wavelength is incident on the metalsurface then the maximum kinetic energy is Kmax = hf1 - hf1 = (hc/1) where1 = 2 or hf1 = (hc/2) or Kmax = (hc/2) - = (1/2) *(hc/) - = (1/2) * hf - or Kmax = (1/2) * 6 - 4 = 3 - 4 = -1eV the negative value of kinetic energy indicates that noelectrons will be emitted. So option (e) is correct.

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