1. Given masses and initial velocities

• Particle $A$: $m_A=\frac{m}{2}, \qquad u_A=v_0$
• Particle $B$: $m_B=\frac{m}{3}, \qquad u_B=0$

The collision is elastic and along the $x$-axis.

2. Use 1D elastic collision formula

For a head-on elastic collision, the final velocity of particle $A$ is

$v_A=\frac{m_A-m_B}{m_A+m_B}u_A + \frac{2m_B}{m_A+m_B}u_B$

Since $u_B=0$,

$v_A=\frac{m_A-m_B}{m_A+m_B}u_A$

Substitute $m_A=\frac{m}{2}$ and $m_B=\frac{m}{3}$:

$v_A=\frac{\frac{m}{2}-\frac{m}{3}}{\frac{m}{2}+\frac{m}{3}}v_0$

$v_A=\frac{\frac{m}{6}}{\frac{5m}{6}}v_0=\frac{1}{5}v_0$

So after collision,

$v_A'=\frac{v_0}{5}$

3. Write de-Broglie wavelength before collision

De-Broglie wavelength is

$\lambda=\frac{h}{p}$

Initially for particle $A$,

$\lambda_0=\frac{h}{m_A v_0}$

4. De-Broglie wavelength after collision

After collision, momentum of $A$ is

$p_A'=m_A v_A'=m_A\left(\frac{v_0}{5}\right)$

Thus new wavelength is

$\lambda' = \frac{h}{m_A(v_0/5)}=5\frac{h}{m_Av_0}=5\lambda_0$

5. Find change in wavelength

$\Delta \lambda = \lambda' - \lambda_0 = 5\lambda_0-\lambda_0=4\lambda_0$

6. Match with options

$\Delta \lambda = 4\lambda_0$

So the correct option is:

D

7. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.