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 :
- A \(\frac{\lambda}{\sqrt{2}}\)
- B \(\frac{\lambda}{2}\)
- C \(2 \lambda\)
- D \(\sqrt{2} \lambda\)
Answer & Solution
Correct Answer
(C) \(2 \lambda\)
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{ h }{\sqrt{2 mE }}\) \(\lambda^{\prime}=\frac{ h }{\sqrt{2 m \left(\frac{ E }{4}\right)}}=\frac{2 h }{\sqrt{2 mE }}=2 \lambda\)
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