Particle A of mass m_{A} = frac{m}{2} moving | vedclass

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(A) Applying the law of conservation of linear momentum:$\frac{m}{2} v_{0} + \frac{m}{3} (0) = \frac{m}{2} v_{A} + \frac{m}{3} v_{B}$$\frac{v_{0}}{2}

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$\Delta \lambda = 4 \lambda_{0}$

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(A) Applying the law of conservation of linear momentum:$\frac{m}{2} v_{0} + \frac{m}{3} (0) = \frac{m}{2} v_{A} + \frac{m}{3} v_{B}$$\frac{v_{0}}{2} = \frac{v_{A}}{2} + \frac{v_{B}}{3} \Rightarrow v_{0} = v_{A} + \frac{2}{3} v_{B} \Rightarrow 3v_{0} = 3v_{A} + 2v_{B} \quad ....(1)$Since the collision is elastic $(e = 1)$:$e = 1 = \frac{v_{B} - v_{A}}{v_{0}} \Rightarrow v_{0} = v_{B} - v_{A} \quad ....(2)$From equation $(2)$,$v_{B} = v_{0} + v_{A}$. Substituting this into equation $(1)$:$3v_{0} = 3v_{A} + 2(v_{0} + v_{A})$$3v_{0} = 3v_{A} + 2v_{0} + 2v_{A}$$v_{0} = 5v_{A} \Rightarrow v_{A} = \frac{v_{0}}{5}$Initial de-Broglie wavelength of particle $A$:$\lambda_{0} = \frac{h}{m_{A} v_{0}} = \frac{h}{(\frac{m}{2}) v_{0}} = \frac{2h}{mv_{0}}$Final de-Broglie wavelength of particle $A$:$\lambda_{f} = \frac{h}{m_{A} v_{A}} = \frac{h}{(\frac{m}{2}) (\frac{v_{0}}{5})} = \frac{10h}{mv_{0}}$Change in de-Broglie wavelength:$\Delta \lambda = \lambda_{f} - \lambda_{0} = \frac{10h}{mv_{0}} - \frac{2h}{mv_{0}} = \frac{8h}{mv_{0}}$Since $\lambda_{0} = \frac{2h}{mv_{0}}$,then $\Delta \lambda = 4 	imes (\frac{2h}{mv_{0}}) = 4 \lambda_{0}$.

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