Particle A of mass $$m_A = \frac{m}{2}$$ moving along the x-axis with velocity $$v_0$$ collides elastically 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 the wavelength of particle A, in terms of its de-Broglie wavelength $$(\lambda_0)$$ before the collision is:

We have particle A of mass $$m_A=\dfrac{m}{2}$$ moving with initial velocity $$u_1=v_0$$ and particle B of mass $$m_B=\dfrac{m}{3}$$ initially at rest, so $$u_2=0$$.

For a perfectly elastic, head-on collision in one dimension, the standard result for the final velocities (derived from simultaneous conservation of linear momentum and kinetic energy) is first written:

$$v_1=\dfrac{m_1-m_2}{m_1+m_2}\,u_1+\dfrac{2m_2}{m_1+m_2}\,u_2,$$

$$v_2=\dfrac{2m_1}{m_1+m_2}\,u_1+\dfrac{m_2-m_1}{m_1+m_2}\,u_2.$$

Substituting $$u_2=0$$ (because B is at rest) and taking $$m_1=m_A=\dfrac{m}{2},\;m_2=m_B=\dfrac{m}{3}$$ we obtain for particle A:

$$v_1=\dfrac{\dfrac{m}{2}-\dfrac{m}{3}}{\dfrac{m}{2}+\dfrac{m}{3}}\,v_0.$$

Now we simplify the numerator and denominator separately:

Numerator: $$\dfrac{m}{2}-\dfrac{m}{3}=\dfrac{3m-2m}{6}=\dfrac{m}{6}.$$

Denominator: $$\dfrac{m}{2}+\dfrac{m}{3}=\dfrac{3m+2m}{6}=\dfrac{5m}{6}.$$

Hence

$$v_1=\dfrac{\dfrac{m}{6}}{\dfrac{5m}{6}}\,v_0=\dfrac{1}{5}\,v_0.$$

So, after collision, particle A moves along the x-axis with speed $$v_1=\dfrac{v_0}{5}.$$

We now turn to the de-Broglie relation which connects momentum and wavelength:

$$\lambda=\dfrac{h}{p},\quad\text{where }p=mv.$$

Initial momentum of particle A:

$$p_0=m_A\,v_0=\left(\dfrac{m}{2}\right)v_0=\dfrac{m v_0}{2}.$$

Therefore its initial de-Broglie wavelength is

$$\lambda_0=\dfrac{h}{p_0}=\dfrac{h}{\dfrac{m v_0}{2}}=\dfrac{2h}{m v_0}.$$

After the collision the momentum of particle A becomes

$$p_1=m_A\,v_1=\left(\dfrac{m}{2}\right)\left(\dfrac{v_0}{5}\right)=\dfrac{m v_0}{10}.$$

So its final wavelength is

$$\lambda_1=\dfrac{h}{p_1}=\dfrac{h}{\dfrac{m v_0}{10}}=\dfrac{10h}{m v_0}.$$

We now express $$\lambda_1$$ in terms of $$\lambda_0$$. Dividing the two expressions gives

$$\dfrac{\lambda_1}{\lambda_0}=\dfrac{\dfrac{10h}{m v_0}}{\dfrac{2h}{m v_0}}=\dfrac{10}{2}=5.$$

Thus $$\lambda_1=5\lambda_0.$$

The change in wavelength is

$$\Delta\lambda=\lambda_1-\lambda_0=5\lambda_0-\lambda_0=4\lambda_0.$$

Hence, the correct answer is Option D.