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A photoelectric surface has a work function of 2.10 eV. Calculate the maximum kinetic energy, in eV, of electrons ejected from this surface by electromagnetic radiation of wavelength 356 nm.

User MUlferts
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The energy of a photon with wavelength λ is given by:

E = hc/λ

where h is Planck's constant and c is the speed of light.

The work function is the minimum energy required to remove an electron from the surface. The maximum kinetic energy of the ejected electron is the difference between the energy of the incident photon and the work function:

KEmax = E - φ

where φ is the work function.

Substituting the given values, we get:

E = hc/λ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(356 x 10^-9 m) = 5.57 x 10^-19 J

φ = 2.10 eV x (1.60 x 10^-19 J/eV) = 3.36 x 10^-19 J

KEmax = E - φ = 5.57 x 10^-19 J - 3.36 x 10^-19 J = 2.21 x 10^-19 J

Converting to electron volts:

KEmax = (2.21 x 10^-19 J)/(1.60 x 10^-19 J/eV) = 1.38 eV

Therefore, the maximum kinetic energy of the ejected electrons is 1.38 eV.
User Milan Halada
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