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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
  • JEE Mains 2023
  • Easy
  • MCQ

The de Broglie wavelength of an electron having kinetic energy \(E\) is \(\lambda\). If the kinetic energy of electron becomes \(\frac{E}{4}\), then its de-Broglie wavelength will be :

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Answer & Solution

Correct Answer

(C) \(2 \lambda\)

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Detailed explanation

\(\lambda=\frac{ h }{\sqrt{2 mE }}\)

\(\lambda^{\prime}=\frac{ h }{\sqrt{2 m \left(\frac{ E }{4}\right)}}=\frac{2 h }{\sqrt{2 mE }}=2 \lambda\)

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