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JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
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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 $$\lambda _A$$. 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 $$\lambda _B$$ = 2$$\lambda _A$$, then the work function of metal B is :
A
1.5eV
B
4eV
C
2eV
D
3eV

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