\(\frac{m}{2} \times V_{0}+\frac{m}{3} \times(0)=\frac{m}{2} V_{A}+\frac{m}{3} V_{B}\)

\(=\frac{V_{0}}{2}=\frac{V_{A}}{2}+\frac{V_{B}}{3}\) \(....(1)\)

since, collision is elastic \(( e =1)\)

\(e =1=\frac{ V _{ B }- V _{ A }}{ V _{0}} \Rightarrow V _{0}= V _{ B }- V _{ A } \ldots(2)\)

On solving \((1) \&(2): V_{A}=\frac{V_{0}}{5}\)

Now, \(De-\)Broglie wavelength of \(A\) before collision :

\(\lambda_{0}=\frac{ h }{ m _{ A } V _{0}}=\frac{ h }{\left(\frac{ m }{2}\right) V _{0}}\)

\(\Rightarrow \lambda_{0}=\frac{2 h }{ mV _{0}}\)

Final \(De-\)Broglie wavelength :

\(\lambda_{ f }=\frac{ h }{ m _{ A } V _{0}}=\frac{ h }{\frac{ m }{2} \times \frac{ V _{ o }}{5}} \Rightarrow \lambda_{ f }=\frac{10 h }{ mV _{ o }}\)

Now \(\Delta \lambda=\lambda_{f}-\lambda_{0}\)

\(\Delta \lambda=\frac{10 h }{ mV _{0}}-\frac{2 h }{ mV _{0}}\)

\(\Rightarrow \Delta \lambda=\frac{8 h }{ mv _{0}} \Rightarrow \Delta \lambda=4 \times \frac{2 h }{ mv _{0}}\)

\(\Rightarrow \Delta \lambda=4 \lambda_{0}\)