Photoelectric effect
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What phenomenon occurs when electromagnetic radiation illuminates a metal surface?
Photoelectric emission. This is when electrons (photoelectrons) are emitted from the metal surface.
Define 'threshold frequency' in the context of the photoelectric effect.
Threshold frequency (f₀) is the minimum frequency of electromagnetic radiation required to cause photoelectric emission from a specific metal surface. Below this frequency, no electrons are emitted, regardless of intensity.
Define 'threshold wavelength' in the context of the photoelectric effect.
Threshold wavelength (λ₀) is the maximum wavelength of electromagnetic radiation that will cause photoemission. Wavelengths longer than λ₀ will not have enough energy to release electrons.
Explain the relationship between photon energy, work function, and photoelectric emission.
For photoelectric emission to occur, the photon energy (hf) must be greater than or equal to the metal's work function (Φ). The excess energy becomes the kinetic energy of the emitted photoelectron.
State the equation that relates photon energy (hf), work function (Φ), and the maximum kinetic energy (KEmax) of a photoelectron.
The equation is hf = Φ + KEmax, where h is Planck's constant, f is the frequency of the incident radiation, Φ is the work function, and KEmax is the maximum kinetic energy of the emitted electron.
Why is the maximum kinetic energy of photoelectrons independent of the intensity of the incident radiation?
The maximum kinetic energy is determined only by the frequency (energy) of individual photons and the work function of the metal. Intensity affects the *number* of photons, not the energy of each photon.
How is photoelectric current related to the intensity of incident radiation?
Photoelectric current is directly proportional to the intensity of the incident radiation. Increased intensity means more photons, leading to more electrons being emitted per unit time, thus a larger current.
A metal has a work function of 3.0 eV. What is the minimum frequency of light needed to eject photoelectrons from it?
Use hf = Φ, where Φ = 3.0 eV = 3.0 * 1.602 x 10⁻¹⁹ J. Therefore, f = Φ/h = (3.0 * 1.602 x 10⁻¹⁹ J) / (6.63 x 10⁻³⁴ Js) ≈ 7.24 x 10¹⁴ Hz.
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