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MHT CET 2025 (23 Apr, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from MHT CET 2025 (23 Apr, Shift 1)

Q32MHT CET 2025 · 23 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

The energy that should be added to an electron to reduce its de-Broglie wavelength from λ\lambda to λ2\frac{\lambda}{2} is nn times the initial energy. The value of ' nn ' is

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Answer: (b)

  1. λ∝1K⇒K∝1λ2\lambda \propto \dfrac{1}{\sqrt{K}} \Rightarrow K \propto \dfrac{1}{\lambda^2}.
  2. λ→λ2⇒K′=4K\lambda \to \dfrac{\lambda}{2} \Rightarrow K' = 4K
  3. Energy added =4K−K=3K= 4K - K = 3K, so n=3n = 3.
  4. So the answer is (b) 3.
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