Threshold frequency and wavelength of electrons in the photoelectric effect

AI Thread Summary
Electrons are emitted from a metal when they absorb energy from light, with the threshold frequency calculated as f0 = 5.08 × 10^14 Hz using the work function of 2.1 eV. The energy of incident light at 500 nm is 2.48 eV, allowing electrons to escape with a maximum kinetic energy of 0.38 eV. To find the wavelength of emitted electrons, the kinetic energy can be used in the equation λ = hc/√(2mK), resulting in an approximate wavelength of 2 nm. The discussion confirms the calculations and explores the relationship between kinetic energy and emitted electron wavelengths. Understanding these principles is crucial for grasping the photoelectric effect.
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Homework Statement



Electrons are emitted from a metal as a consequence of their absorption of energy from a light beam. Find the threshold frequency of the metal and the wavelength of emitted electrons.

Wavelength of incident light λ = 500 nm
Work function of the metal \phi = 2.1 eV

Homework Equations



Threshold frequency:

f0 = \frac{\phi}{h}

The Attempt at a Solution



Threshold frequency:

f0 = \frac{2.1 eV}{4.136 \times 10^{-15} eV·s } = 5.08 × 1014 Hz

Is this correct?

How can I find the wavelength of emitted electrons?
 
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Thanks for the confirmation.

As I understand the photoelectric effect, one photon transfers all of its energy to an electron, so the energy absorbed by any electron must be the same (for a monochromatic beam), in this case hc/(500 nm) = 2.48 eV. Electrons need at least 2.1 eV to escape this metal, so they must have a maximum kinetic energy of 0.38 eV. Since not all of them have the same kinetic energy (due to losses) I assume they must have different wavelengths as well, how can I find the wavelength of emitted electrons?
 
You can assume that they all have 0.38 eV. If you like, use "<=" in the calculations, but that won't change much.
 
Ok, so it would be λ = \frac{hc}{\sqrt{2mc^2K}} \approx 2 nm, right?
 
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