ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter

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 :

  1. A \(\frac{\lambda}{\sqrt{2}}\)
  2. B \(\frac{\lambda}{2}\)
  3. C \(2 \lambda\)
  4. D \(\sqrt{2} \lambda\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2 \lambda\)

Step-by-step Solution

Detailed explanation

\(\lambda=\frac{ h }{\sqrt{2 mE }}\) \(\lambda^{\prime}=\frac{ h }{\sqrt{2 m \left(\frac{ E }{4}\right)}}=\frac{2 h }{\sqrt{2 mE }}=2 \lambda\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app