The de Broglie wavelength associated with a moving particle $A$ is precisely $4.0 imes 10^{-10} \m

Question

The de Broglie wavelength associated with a moving particle $A$ is precisely $4.0 	imes 10^{-10} \mathrm{~m}$. Another particle $B$, having the exact same mass as $A$, possesses a kinetic energy that is strictly 4 times that of $A$. The de Broglie wavelength associated with particle $B$ is

Professor PC Thomas Classes · Answer

(a) : The de Broglie wavelength related to kinetic energy is expressed as $\lambda = \frac{h}{\sqrt{2m(KE)}}$. Given that the rest mass $m$ is entirely identical for both particles, we can deduce $\lambda \propto \frac{1}{\sqrt{KE}}$. Because the kinetic energy of particle $B$ is 4 times larger than that of $A$, its resulting wavelength drops to $\lambda_B = \frac{\lambda_A}{\sqrt{4}} = \frac{\lambda_A}{2}$. Consequently, $\lambda_B = \frac{4.0 	imes 10^{-10}}{2} = 2.0 	imes 10^{-10} \mathrm{~m}$.