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JEE PHYSICS MCQ: DUAL NATURE OF RADIATION AND MATTER

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Analyzing +4 | -1
When photon of energy 4.0  eV4.0\,\,eV strikes the surface of a metal A,A, the ejected photoelectrons have maximum kinetic energy TA eV{T_A}\,eV and de-Broglie wavelength λA.{\lambda _A}. The maximum kinetic energy of photoelectrons liberated from another metal BB by photon of energy 4.50  eV4.50\,\,eV is TB=(TA−1.5) eV.{T_B} = \left( {{T_A} - 1.5} \right)\,eV. If the de-Broglie wavelength of these photoelectrons λB=2λA,{\lambda _B} = 2{\lambda _A}, then the work function of metal BB is
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Correct option: A
4 eV4\,eV
Option A ($4 eV$) is correct because by using the photoelectric effect equation and the de-Broglie wavelength relation, we can find the work function of metal B.

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