When photons of energy $$4.25$$ eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy TA and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy $$4.70$$ eV is TB $$=$$ $$TA-1.50$$ eV. If the de Broglie wavelength of these photoelectrons is λ$$B=2$$λA, then

Question

When photons of energy $$4.25$$ eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy TA and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy $$4.70$$ eV is TB $$=$$ $$TA-1.50$$ eV. If the de Broglie wavelength of these photoelectrons is λ$$B=2$$λA, then

Infinity Learn · Accepted Answer

For metal A:$$4.25=WA+TA$$  ……..(i) Also, $$TA=12mvA2=12m2vA2m=pA22m=h22m$$λ$$A2$$   ………(ii)                                                                          ∵λ$$=hp$$ For metal B:$$4.7=TA$$−$$1.5+WB$$    ………(iii) Also, $$TB=h22m$$λ$$B2$$×$$2m$$λ$$A2h2=$$λ$$A2$$λ$$B2$$⇒TA−$$1.5TA=$$λ$$A22$$λ$$A2=$$λ$$A24$$λ$$A2=14$$ ∵λ$$B=2$$λA given ⇒$$4TA$$−$$6=TA$$⇒$$TA=2eV$$∴$$WA=2.25eV$$,$$WB=4.20eV$$,$$TB=(2-1.5)eV$$ $$=$$ $$0.5$$ ev

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