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Question: Answered & Verified by Expert
The work function of the metal A is equal to the ionization energy of the hydrogen atom in the first excited state. The work function of the metal B is equal to the ionization energy of He+ ion in the second orbit. Photons of the same energy E are incident on both A and B. The maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B. Value of E ( in eV) is
PhysicsDual Nature of MatterNEET
Options:
  • A 23.8 eV
  • B 20.8 eV
  • C 32.2 eV
  • D 24.6 eV
Solution:
2345 Upvotes Verified Answer
The correct answer is: 23.8 eV
En=13.6Z2n2

WA= ionization energy of an electron in 2nd orbit of the hydrogen atom = 3.4 eV

WB = ionization energy of an electron in 2nd orbit of He+ ion

      = 13.6 eV.

Now, given that

KA=2KB

or   E-WA=2E-WB

∴  E=2WB-WA=23.8 eV

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