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Question: Answered & Verified by Expert
Photoelectrons are emitted when 4000  radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region having α -particles to form He+ ion, emitting a single photon. In this process, He+ ions thus formed are in their fourth excited state. The energy released during the combination of He+ ions is
PhysicsAtomic PhysicsJEE Main
Options:
  • A 5.38 eV
     
  • B 3.38 eV
     
  • C 2.38 eV
     
  • D 1.38 eV
     
Solution:
2860 Upvotes Verified Answer
The correct answer is: 3.38 eV
 
The total energy of electron initially = hc λ w=1.18

Final total energy =2.2

The difference of energy when photoelectron recombines with α -particle

=1.18--2.2

=3.38 eV

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