If λ be the de-Broglie wavelength of a thermal neutron at 27°C. The wavelength of the same neutron a — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - de Broglie Wavelength of a Thermal Neutron The de Broglie wavelength of a particle with kinetic energy $K.E.$ is: $$\lambda = \frac{h}{\sqrt{2m(K.E.)}}$$ For a thermal neutron in equilibrium at absolute temperature $T$: $$K.E. = \frac{3}{2}k_B T$$ Substituting: $$\lambda = \frac{h}{\sqrt{3mk_B T}}$$ Since $h$, $m$, $k_B$ are constants: $$\lambda \propto \frac{1}{\sqrt{T}}$$ ### Step 2 - Compare Wavelengths at Two Temperatures $$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}}$$ ### Step 3 - Convert Temperatures to Kelvin and Calculate $$T_1 = 27 + 273 = 300\text{ K}$$ $$T_2 = 927 + 273 = 1200\text{ K}$$ $$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2} = 0.5$$ $$\lambda_2 = \boxed{0.5\lambda}$$ ### Step 4 - Analysis of Options * **Option (A) $\lambda$:** Incorrect — wavelength changes with temperature. * **Option (B) $0.5\lambda$:** Correct — temperature quadrupled (300K to 1200K), so wavelength halved. * **Option (C) $2\lambda$:** Incorrect — would require temperature to decrease to 75 K. * **Option (D) $0.25\lambda$:** Incorrect — would apply if $\lambda \propto 1/T$ instead of $1/\sqrt{T}$.