# Solve this following

Question:

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 :

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

2. $\Delta \lambda=\frac{5}{2} \lambda_{0}$

3. $\Delta \lambda=2 \lambda_{0}$

4. $\Delta \lambda=\frac{3}{2} \lambda_{0}$

Correct Option: 1

Solution:

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.