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

💡 Solution & Explanation

The de-Broglie wavelength of a thermal particle is $\lambda = \frac{h}{\sqrt{3mkT}}$, so $\lambda \propto \frac{1}{\sqrt{T}}$. Here $T_1 = 27 + 273 = 300$ K and $T_2 = 927 + 273 = 1200$ K. $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = 0.5$. Thus, $\lambda_2 = 0.5\lambda$. Therefore, correct answer is B.