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Question: Answered & Verified by Expert
The kinetic energy of an electron having de-Broglie wavelength λ is ( h= Planck's constant, m= mass of electron)
PhysicsAtomic PhysicsJEE Main
Options:
  • A h2mλ
  • B h22mλ2
  • C h22m2λ2
  • D h22m2λ
Solution:
1515 Upvotes Verified Answer
The correct answer is: h22mλ2
λ= h 2mK.E

λ 2 = h 2 2m( K.E )

( K.E )= h 2 2m λ 2
 

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