Two particles A and B are in motion. If the de Broglie wavelength associated with particle A is and — Atomic Structure Chemistry Question

💡 Solution & Explanation

According to the de Broglie equation, wavelength $\lambda = \frac{h}{p}$, which means $\lambda \propto \frac{1}{p}$. Since the momentum of particle B is half that of A ($p_B = p_A / 2$), its wavelength will be twice that of A. $\lambda_B = 2 \lambda_A = 2 \times 5 \times 10^{-8} \text{ m} = 10 \times 10^{-8} \text{ m} = 10^{-7} \text{ m}$.