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NEET UG CHEMISTRY MCQ: STRUCTURE OF ATOM

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Analyzing +4 | -1
When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA{T_A} (expressed in eV) and de Broglie wavelength λA.{\lambda _A}. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70V is TB=TA−1.50  eV.{T_B} = {T_A} - 1.50\;\mathrm{eV}. If the de Broglie wavelength of these photoelectrons is λB=2λA,{\lambda _B} = 2{\lambda _A}, then which is not correct?
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Correct option: B
The work function of B is 3.70 eV
Option B (The work function of B is 3.70 eV) is not correct. The work function of B is calculated to be 2.75 eV, not 3.70 eV.

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