The given figure shows the variation of de Broglie wavelength ( λ )...

The de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by:

$\lambda = \dfrac{h}{\sqrt{2mqV}}$

This can be rearranged as:

$\lambda = \left(\dfrac{h}{\sqrt{2mq}}\right) \dfrac{1}{\sqrt{V}}$

The graph of $\lambda$ versus $\dfrac{1}{\sqrt{V}}$ is a straight line passing through the origin with slope $S = \dfrac{h}{\sqrt{2mq}}$.

For a proton, let the mass be $m_p = m$ and charge be $q_p = e$. The slope is:

$S_p = \dfrac{h}{\sqrt{2me}}$

For an $\alpha$-particle, the mass is $m_\alpha = 4m$ and charge is $q_\alpha = 2e$. The slope is:

$S_\alpha = \dfrac{h}{\sqrt{2(4m)(2e)}} = \dfrac{h}{\sqrt{16me}} = \dfrac{1}{2\sqrt{2}} \dfrac{h}{\sqrt{2me}} = \dfrac{S_p}{2\sqrt{2}}$

Since $S_p > S_\alpha$, the line corresponding to the proton must have a greater slope than the line corresponding to the $\alpha$-particle.

From the given graph, Line A has a greater slope than Line B. Therefore, Line A represents the proton and Line B represents the $\alpha$-particle.

Answer: Line A represents the proton and Line B represents the $\alpha$-particle.