The de-Broglie wavelength of a vehicle moving with velocity is . Its load is changed so that the vel — Atomic Structure Chemistry Question

💡 Solution & Explanation

Initial kinetic energy $K = \frac{1}{2}mv^2$. New velocity is $2v$, new K.E. is $2K$. So $2K = \frac{1}{2}m_{new}(2v)^2 = 2m_{new}v^2$. Since $2\left(\frac{1}{2}mv^2\right) = 2m_{new}v^2$, we find $m_{new} = m/2$. The new momentum $p_{new} = m_{new} \times 2v = (m/2)(2v) = mv = p$. Since momentum is unchanged, wavelength $\lambda$ is unchanged. Therefore, correct answer is A.