Determine the wavelength of light (in nm) irradiating a photocathode with a work function of A = 3.7 eV, if the maximum kinetic energy of photoelectrons is T max = 1.7 eV.
Av = 3.7 eV = 5.92 * 10 ^ -19 J.
Tmax = 1.7 eV = 2.72 * 10 ^ -19 J.
h = 6.6 * 10 ^ -34 J * s.
C = 3 * 10 ^ 8 m / s.
The energy of the incident photons Eph goes to knock out electrons from the surface of the metal AB and impart kinetic energy Tmax to them.
Eph = Av + Tmax.
The energy of photons Eph of light that fall on the photocathode can be expressed by the formula: Ef = h * C / λ, where h is Planck’s constant, C is the speed of light, λ is the wavelength of photons.
h * C / λ = Av + Tmax.
λ = h * C / (Av + Tmax).
λ = 6.6 * 10 ^ -34 J * s * 3 * 10 ^ 8 m / s / (5.92 * 10 ^ -19 J + 2.72 * 10 ^ -19 J) = 25.46 * 10 ^ -9 m = 25.46 nm.
Answer: the photocathode is irradiated with a wavelength λ = 25.46 nm.
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