A particle $\text{A}$ of mass $m_A=\frac{m}{2}$ moving along the $x-$axis with velocity $v_0$ collides elastically with another particle $\text{B}$ at rest having mass $m_B=\frac{m}{3}$. If both particle move along the $x-$ axis after the collision. The change $\Delta \lambda$ in de-Broglie wavelength of particle $\text{A}$, in terms of its de-Broglie wavelength $\left({\lambda }_0\right)$ before collision is-

The correct option is D $\Delta \lambda =4{\lambda }_0$

Applying momentum conservation,

$\frac{m}{2}\times V_0+\frac{m}{3}\times \left(0\right)=\frac{m}{2}{V}_A+\frac{m}{3}{V}_B$

$\frac{V_0}{2}=\frac{V_A}{2}+\frac{V_B}{3}....\left(1\right)$

Since, collision is elastic, $e=1$

$e=\frac{V_B-V_A}{u_A-u_B}[\because u_A=V_o;u_B=0]$

$\Rightarrow V_0=V_B-V_A....\left(2\right)$
On solving equations $\left(1\right)$ and $\left(2\right)$ we get,

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

Now, de-Broglie wavelength of $\text{A}$ before collision is,

${\lambda }_0=\frac{h}{m_A{V}_0}=\frac{h}{\left(\frac{m}{2}\right)V_0}$

$\Rightarrow {\lambda }_0=\frac{2h}{m{V}_0}$

The final de-Broglie wavelength is,

${\lambda }^{{}^{\prime }}=\frac{h}{m_A{V}_A^{{}^{\prime }}}=\frac{h}{\frac{m}{2}\times \frac{V_0}{5}}=\frac{10h}{m{V}_0}$

$\therefore \Delta \lambda ={\lambda }^{{}^{\prime }}-{\lambda }_0=\frac{10h}{m{V}_0}-\frac{2h}{m{V}_0}=\frac{8h}{m{V}_0}$

$\Rightarrow \Delta \lambda =4\times \frac{2h}{m{v}_0}=4{\lambda }_0$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.