The de-Broglie wavelength of a neutron at 27 is ' λ₀ '. What will be its wavelength at 927? (a) (λ₀)/(4) (b) (λ₀)/(3) (c) (λ₀)/(2) (d) (3λ₀)/(2)

Answer: (c). λ = (h)/(√(3mkT)) ⇒ λ ∝ 1/√(T). T₁ = 300 K, T₂ = 1200 K, so T₂ = 4T₁. λ₂ = (λ₀)/(√(4)) = (λ₀)/(2) So the answer is (c) (λ₀)/(2).

The de-Broglie wavelength of a neutron at 27 is ' λ₀ '. What will be… | MHT CET 2025 (22 Apr, Shift 2)

Q34MHT CET 2025 · 22 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

The de-Broglie wavelength of a neutron at $27^\circ\text{C}$ is ' $\lambda_0$ '. What will be its wavelength at $927^\circ\text{C}$ ?

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Answer: (c)

$\lambda = \dfrac{h}{\sqrt{3mkT}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{T}}$ .
$T_1 = 300\text{ K}$ , $T_2 = 1200\text{ K}$ , so $T_2 = 4T_1$ .
$\lambda_2 = \dfrac{\lambda_0}{\sqrt{4}} = \dfrac{\lambda_0}{2}$
So the answer is (c) $\frac{\lambda_0}{2}$ .