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Calculate the kinetic energy of a photoelectron (in eV) emitted on shining light of wavelength 6.2 × 10-6 m on a metal surface. The work function of the metal is 0.1 eV.
  • a)
    10 eV
  • b)
    0.1 eV
  • c)
    0.01 eV
  • d)
    1000 eV
Correct answer is option 'B'. Can you explain this answer?
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Given data:
- Wavelength of light (λ) = 6.2 × 10^-6 m
- Work function of the metal (φ) = 0.1 eV

Calculate the energy of a photon:
- We know that energy of a photon is given by E = hc/λ, where h is the Planck's constant (6.626 × 10^-34 J s) and c is the speed of light (3 × 10^8 m/s).
- Converting the wavelength to meters: 6.2 × 10^-6 m = 6.2 × 10^-9 m
- Energy of a photon = (6.626 × 10^-34 J s * 3 × 10^8 m/s) / (6.2 × 10^-9 m) ≈ 3.2 × 10^-19 J
- Converting the energy from Joules to electron volts (1 eV = 1.6 × 10^-19 J), we get: 3.2 × 10^-19 J / 1.6 × 10^-19 J/eV = 2 eV

Calculate the kinetic energy of the photoelectron:
- The kinetic energy of the photoelectron is given by the difference between the energy of the photon and the work function of the metal.
- Kinetic energy = Energy of photon - Work function = 2 eV - 0.1 eV = 1.9 eV
Therefore, the kinetic energy of the photoelectron emitted on shining light of wavelength 6.2 × 10^-6 m on a metal surface is 1.9 eV, which is closest to option B (0.1 eV).
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Calculate the kinetic energy of a photoelectron (in eV) emitted on shining light of wavelength 6.2 × 10-6 m on a metal surface. The work function of the metal is 0.1 eV.a)10 eVb)0.1 eVc)0.01 eVd)1000 eVCorrect answer is option 'B'. Can you explain this answer?
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