The energy that should be added to an electron to reduce its de-Broglie wavelength from λ to (λ)/(2) is n times the initial energy. The value of ' n ' is (a) 4 (b) 3 (c) 2 (d) 1

Answer: (b). λ ∝ 1/√(K) ⇒ K ∝ (1)/(λ²). λ → (λ)/(2) ⇒ K' = 4K Energy added = 4K - K = 3K, so n = 3. So the answer is (b) 3.

The energy that should be added to an electron to reduce its… | 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 $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of ' $n$ ' is

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Answer: (b)

$\lambda \propto \dfrac{1}{\sqrt{K}} \Rightarrow K \propto \dfrac{1}{\lambda^2}$ .
$\lambda \to \dfrac{\lambda}{2} \Rightarrow K' = 4K$
Energy added $= 4K - K = 3K$ , so $n = 3$ .
So the answer is (b) 3.

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