When photons of energy 4.25 eV strike the surface of metal A , the ejected photoelectrons…
Chemistry · JEE Advanced · NTA Exams — Atomic Structure
When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA eV 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 λB = 2λA, then
- the work function of A is 2.25 eV
- the work function of B is 4.20 eV
- TA = 2.00 eV
- TB = 2.75 eV.
Answer
(C) T A = 2.00 eV
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