Particle A of mass m_A = ${m \over 2}$ moving along the x-axis with velocity v₀ collides elastically with another particle B at rest having mass m_B = ${m \over 3}$. If both particles move along the x-axis after the collision, the change $\Delta$$\lambda$ in de-Broglie wavlength of particle A, in terms of its de-Broglie wavelength ($\lambda$₀) before collision is :

Applying momentum conservation <br><br>$${m \over 2} \times {V_0} + {m \over 3} \times 0 = {m \over 2}{V_A} + {m \over 3}{V_B}$$ <br><br>$\Rightarrow$ ${{{V_0}} \over 2} = {{{V_A}} \over 2} + {{{V_B}} \over 3}$ .....(1) <br><br>Since, collision is elastic (e = 1) <br><br>e = 1 = ${{{V_B} - {V_A}} \over {{V_0}}}$ <br><br>$\Rightarrow$ V₀ = V_B – V_A .....(2) <br><br>On solving (1) &amp; (2) : <br><br>V_A = ${{{V_0}} \over 5}$ <br><br>Now, De-Broglie wavelength of A before collision : <br><br>$\lambda$₀ = ${h \over {{m_A}{V_0}}}$ <br><br>= ${h \over {\left( {{m \over 2}} \right){V_0}}}$ <br><br>= ${{2h} \over {m{V_0}}}$ <br><br>Final De-Broglie wavelength : <br><br>$\lambda$_f = ${h \over {{m_A}{V_A}}}$ <br><br>= ${h \over {\left( {{m \over 2}} \right)\left( {{{{V_0}} \over 5}} \right)}}$ = ${{10h} \over {m{V_0}}}$ <br><br>Now, $\Delta$$\lambda$ = $\lambda$_f - $\lambda$₀ <br><br>= ${{10h} \over {m{V_0}}}$ - ${{2h} \over {m{V_0}}}$ <br><br>$\Rightarrow$ $\Delta$$\lambda$ = ${{8h} \over {m{V_0}}}$ <br><br>$\Rightarrow$ $\Delta$$\lambda$ = $4 \times {{2h} \over {m{V_0}}}$ <br><br>$\Rightarrow$ $\Delta$$\lambda$ = 4$\lambda$₀