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Sagot :
Aluminium has a work function of 4. 08 eV. The cutoff wavelength and cutoff frequency for the photoelectric effect is 303.9nm and 911.7× 10¹⁷ s⁻¹ respectively.
Work Function is the minimum energy required to eject an electron from a photoelectric material.
Cutoff Wavelength is the maximum wavelength below which electron will be ejected
Cutoff Frequency is the minimum frequency required to eject electron.
Let the work function of Aluminium be Φ
Given, work function Φ = 4.08eV
We know that hc/λ = Φ
where, h is Planck constant
c is speed of light
λ is wavelength of light used
Hence, on subsitution
1240 / λ = 4.08 eV (hc = 1240)
⇒ λ = 303.9 nm
Hence, cutoff wavelength used is 303.9 nm
We know that ν = c/λ
ν = 3 × 10⁸ / 303.9
⇒ ν = 911.7 × 10¹⁷ s⁻¹
Hence, cutoff frequency used is 911.7 × 10¹⁷ s⁻¹
Learn more about Work function here, https://brainly.com/question/24180170
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