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A photon of wavelength 124 nm is absorbed by a metal whose work function is 6 eV. The maximum uncertainty in the momentum of the ejected photoelectron is (h/(4*pi)) * 10⁹ kg*m/s. Given that (6.135)² / pi = 12, identify the correct statements.
- Maximum kinetic energy of the ejected photoelectron is 4 eV.
- Maximum uncertainty in position of the electron is 1 nm.
- Maximum uncertainty in de-Broglie wavelength is 3 pm.
- Maximum uncertainty in de-Broglie wavelength is approximately 30 pm.
Correct answer: Maximum kinetic energy of the ejected photoelectron is 4 eV.
Solution
Photon energy is 10 eV, work function is 6 eV, so KE_max = 4 eV. Using Heisenberg's principle with the given deltaₚ, deltaₓ = 1 nm. The uncertainty in de-Broglie wavelength can be computed from delta_lambda/lambda = deltaₚ/p.
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