#### Particle A of mass $m_{A}=\frac{m}{2}$ moving along the axis with velocity $v_{0}$ collides eleastically with another particle B at rest having mass $m_{B}=\frac{m}{3}$. If both particles move along the x-axis after the collision, the change $\Delta \lambda$ in de-Brogile wavelength of particle A, in terms of its de-Brogile wavelength $(\lambda _{0})$ before collision is: Option: 1 Option: 2 Option: 3 Option: 4

Applying momentum conservation
$\frac{\mathrm{m}}{2} \times \mathrm{V}_{0}+\frac{\mathrm{m}}{3} \times(0)=\frac{\mathrm{m}}{2} \mathrm{~V}_{\mathrm{A}}+\frac{\mathrm{m}}{3} \mathrm{~V}_{\mathrm{B}} \\ \Rightarrow \frac{\mathrm{V}_{0}}{2}=\frac{\mathrm{V}_{\mathrm{A}}}{2}+\frac{\mathrm{V}_{\mathrm{B}}}{3} \quad \ldots (1)$
Since, collision is elastic (e=1)
$\mathrm{e}=1=\frac{\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{0}} \\ \Rightarrow \mathrm{V}_{0}=\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}} \ldots (2)$

From equation (1) and (2)

we get

$V_A=\frac{V_0}{5}$

Now, De-Broglie wavelength of A before collision:

$\begin{array}{l} \lambda_{0}=\frac{h}{m_{A} V_{0}}=\frac{h}{\left(\frac{m}{2}\right) V_{0}} \\ \\ \Rightarrow \lambda_{0}=\frac{2 h}{m V_{0}} \end{array}$

Final De-Broglie wavelength
$\lambda_{\mathrm{f}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{V}_{0}}=\frac{\mathrm{h}}{\frac{\mathrm{m}}{2} \times \frac{\mathrm{V}_{0}}{5}} \\ \\ \Rightarrow \lambda_{\mathrm{f}}=\frac{10 \mathrm{~h}}{\mathrm{mV}_{0}}$

$\begin{array}{l} \text { Now } \Delta \lambda=\lambda_{\mathrm{f}}-\lambda_{0} \\ \Delta \lambda=\frac{10 \mathrm{~h}}{\mathrm{mV}_{0}}-\frac{2 \mathrm{~h}}{\mathrm{mV}_{0}} \\ \Rightarrow \Delta \lambda=\frac{8 \mathrm{~h}}{\mathrm{mv}_{0}} \\\Rightarrow \Delta \lambda=4 \times \frac{2 \mathrm{~h}}{\mathrm{mv}_{0}} \\ \Rightarrow \Delta \lambda=4 \lambda_{0} \end{array}$