Particle $\mathrm{A}$ of mass $\mathrm{m}_{\mathrm{A}}=\frac{\mathrm{m}}{2}$ moving along the
$x$-axis with velocity $v_{0}$ collides elastically with
another particle $\mathrm{B}$ at rest having mass $\mathrm{m}_{\mathrm{B}}=\frac{\mathrm{m}}{3}$.
If both particles move along the $x$-axis after the collision, the change $\Delta \lambda$ in de-Broglie wavelength of particle A, in terms of its de-Broglie wavelength $\left(\lambda_{0}\right)$ before collision is :
Correct Option: 1
Applying momentum conservation
$\frac{\mathrm{m}}{2} \times \mathrm{V}_{0}+\frac{\mathrm{m}}{3} \times(0)=\frac{\mathrm{m}}{2} \mathrm{~V}_{\mathrm{A}}+\frac{\mathrm{m}}{3} \mathrm{~V}_{\mathrm{B}}$
$=\frac{\mathrm{V}_{0}}{2}=\frac{\mathrm{V}_{\mathrm{A}}}{2}+\frac{\mathrm{V}_{\mathrm{B}}}{3}$ ................(1)
Since, collision is elastic $(\mathrm{e}=1)$
$\mathrm{e}=1=\frac{\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{0}} \Rightarrow \mathrm{V}_{0}=\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}} \cdots$ (2)
On solving (1) \& (2) : $\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{V}_{0}}{5}$
Now, De-Broglie wavelength of A before collision :
$\lambda_{0}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{V}_{0}}=\frac{\mathrm{h}}{\left(\frac{\mathrm{m}}{2}\right) \mathrm{V}_{0}}$
$\Rightarrow \lambda_{0}=\frac{2 \mathrm{~h}}{\mathrm{mV}_{0}}$
Final De-Broglie wavelength :
$\lambda_{\mathrm{f}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{V}_{0}}=\frac{\mathrm{h}}{\frac{\mathrm{m}}{2} \times \frac{\mathrm{V}_{0}}{5}} \Rightarrow \lambda_{\mathrm{f}}=\frac{10 \mathrm{~h}}{\mathrm{mV}_{0}}$
Now $\Delta \lambda=\lambda_{\mathrm{f}}-\lambda_{0}$
$\Delta \lambda=\frac{10 \mathrm{~h}}{\mathrm{mV}_{0}}-\frac{2 \mathrm{~h}}{\mathrm{mV}_{0}}$
$\Rightarrow \Delta \lambda=\frac{8 h}{m v_{0}} \Rightarrow \Delta \lambda=4 \times \frac{2 h}{m v_{0}}$
$\Rightarrow \Delta \lambda=4 \lambda_{0}$
option (1) is correct.