JEE Main 8 Jan 2020 Shift 1 Solved Paper

Section: Physics

Question : 2 of 75

Marks: +1, -0

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

[8 Jan 2020 Shift 1]

3 eV
2 eV
4 eV
1.5 ev

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