A particle $$A$$ of mass $$m$$ and initial velocity $$v$$ collides with a particle $$B$$ of mass $$\frac{m}{2}$$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $$\lambda_{A}$$ to $$\lambda_{B}$$ after the collision is:

We begin by noting the data given. Particle $$A$$ has mass $$m$$ and initial speed $$v$$, while particle $$B$$ has mass $$\dfrac{m}{2}$$ and is initially at rest. The collision is one-dimensional, perfectly elastic, and therefore both linear momentum and kinetic energy remain conserved.

For a head-on elastic collision in one dimension, the standard kinematic relations for the final velocities are first written. If the initial velocities are $$u_1$$ and $$u_2$$ for masses $$m_1$$ and $$m_2$$ respectively, then

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

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

Here we identify $$m_1=m$$, $$u_1=v$$, $$m_2=\dfrac{m}{2}$$ and $$u_2=0$$ (because particle $$B$$ is initially at rest). We now substitute these quantities one at a time.

First, the common denominator that will appear is

$$m_1+m_2=m+\frac{m}{2}=\frac{3m}{2}.$$

Now for the final speed of particle $$A$$ (call it $$v_A$$):

$$v_A=v_1=\frac{m_1-m_2}{m_1+m_2}\,u_1 =\frac{m-\dfrac{m}{2}}{\dfrac{3m}{2}}\;v =\frac{\dfrac{m}{2}}{\dfrac{3m}{2}}\;v =\frac{m}{2}\times\frac{2}{3m}\;v =\frac{1}{3}\,v.$$

In exactly the same way, the final speed of particle $$B$$ (call it $$v_B$$) is

$$v_B=v_2=\frac{2m_1}{m_1+m_2}\,u_1 =\frac{2m}{\dfrac{3m}{2}}\;v =2m\times\frac{2}{3m}\;v =\frac{4}{3}\,v.$$

The numerical values we have obtained are therefore

$$v_A=\frac{v}{3}, \qquad v_B=\frac{4v}{3}.$$

We now switch to de-Broglie wavelengths. The fundamental relation is always stated explicitly:

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

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum $$p=mv$$ of the particle.

For particle $$A$$ after the collision, the momentum is

$$p_A=m\,v_A=m\left(\frac{v}{3}\right)=\frac{mv}{3}.$$

Thus its de-Broglie wavelength is

$$\lambda_A=\frac{h}{p_A}=\frac{h}{\dfrac{mv}{3}}=\frac{3h}{mv}.$$

For particle $$B$$ after the collision, the momentum is

$$p_B=\left(\frac{m}{2}\right)\,v_B=\left(\frac{m}{2}\right)\left(\frac{4v}{3}\right) =\frac{2mv}{3}.$$

So its de-Broglie wavelength is

$$\lambda_B=\frac{h}{p_B}=\frac{h}{\dfrac{2mv}{3}}=\frac{3h}{2mv}.$$

To obtain the required ratio, we divide $$\lambda_A$$ by $$\lambda_B$$ step by step:

$$\frac{\lambda_A}{\lambda_B} =\frac{\dfrac{3h}{mv}}{\dfrac{3h}{2mv}} =\frac{3h}{mv}\times\frac{2mv}{3h} =2.$$

All factors $$3h$$ and $$mv$$ cancel exactly, leaving the numerical value $$2$$. This ratio matches Option C in the list.

Hence, the correct answer is Option C.