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mhooker_10
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What frequency of light must be used to eject electrons from a gold surface with a maximum kinetic energy of 6.48×10(−19)J ?
The work function of gold is 4.58 eV.
f=
The work function of gold is 4.58 eV.
f=
The frequency of light required to eject electrons from gold is approximately 5 x 10^14 Hz. This frequency falls in the ultraviolet range of the electromagnetic spectrum.
The frequency of light determines the energy of each photon, which in turn affects the amount of energy transferred to the electrons in the gold. If the frequency is too low, the electrons will not have enough energy to be ejected. If the frequency is too high, the electrons will be ejected with excess energy, potentially causing damage to the gold surface.
No, only light with a frequency higher than the threshold frequency of gold can eject electrons. This is known as the photoelectric effect, and it follows the principle that the energy of a photon must be greater than the work function of the metal in order for electrons to be emitted.
Yes, the intensity of light does affect the ejection of electrons from gold. The higher the intensity, the more photons are present, and therefore the more electrons will be ejected. However, the intensity does not affect the energy of each individual photon.
The energy of ejected electrons is directly proportional to the frequency of light. This means that as the frequency increases, so does the energy of the ejected electrons. This relationship is described by the equation E = hf, where E is the energy of the electron, h is Planck's constant, and f is the frequency of light.