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PhysicsMediumMCQ2020 · 08 Jan Shift 1

Q18.When photon of energy 4.0eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAeV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is TB = (TA −1.5)eV. If the de-Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is: (1) 4eV (2) 2eV (3) 1.5eV (4) 3eV

What This Question Tests

This problem combines the photoelectric effect with the de-Broglie wavelength concept, requiring setting up and solving equations for kinetic energy and wavelength for two different metals.

Concepts Tested

Einstein's photoelectric equationde-Broglie wavelength for electronKinetic energy relation

Formulas Used

KE_max = hν - Φ

λ = h/√(2mKE_max)

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