When the light of wavelength is falling on gold surface, the electrons begin to exchange energies.
a)The work function in eV is Φ =5.097 eV.
b) The cut-off wavelength is λ₀ = 243.71 nm
c) The frequency is ν₀ =1.231 × 10¹⁵ Hz
<h3>What is work function?</h3>
The energy needed for a particle to escape and break through the surface.
The kinetic energy of the light emitted is 2.66 eV and wavelength of the light is 160 nm = 160 × 10⁻⁹ m.
a) The work function of the gold for given maximum kinetic energy is
Φ = hc / λ - K.Emax
Substituting 6.626 × 10⁻³⁴ J.s for h, 3 × 10⁸ m/s for c and 2.66 eV for K.Emax, work function will be
Φ =8.16 × 10⁻¹⁹ J
1 eV = 1.6 × 10⁻¹⁹
The work function in eV is Φ =5.097 eV.
b) The cutoff wavelength is related to work function as
λ₀ = hc / Φ
Substitute the corresponding values into the equation, we get the cut off wavelength
λ₀ = 243.71 nm
c) The frequency corresponding to the cut-off wavelength is
ν₀ = c / λ₀
Substitute the corresponding values into the equation, we get the frequency,
ν₀ =1.231 × 10¹⁵ Hz
Therefore, the values for the following are
a)The work function in eV is Φ =5.097 eV.
b) The cut-off wavelength is λ₀ = 243.71 nm
c) The frequency is ν₀ =1.231 × 10¹⁵ Hz
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