When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons…
Chemistry · JEE Main · NTA Exams — Atomic Structure
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 λA. 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 λB = 2λA, then the work function of metal B is:
- 3 eV
- 2 eV
- 4 eV
- 1.5 eV
Answer
(A) 3 eV
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