A proton accelerated through a potential difference of V volts has a de-Broglie wavelength λ associated with it. In order to get the same wavelength associated with an α-particle, the required accelerating potential is

The correct option is V/8.

Let us understand the step-by-step physical and mathematical reasoning to arrive at this answer.

The de-Broglie wavelength ($\lambda$) associated with a particle of mass $m$ and charge $q$ accelerated from rest through a potential difference $V$ is given by the formula:
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$
where $h$ is Planck's constant, $p$ is the momentum, and $K$ is the kinetic energy of the particle.

Since the kinetic energy gained by a charged particle accelerated through a potential difference $V$ is $K=qV$, the de-Broglie wavelength can be written as:
$\lambda =\frac{h}{\sqrt{2mqV}}$

Let us compare a proton (denoted by subscript $p$) and an α-particle (denoted by subscript $\alpha$):
1. A proton has mass $m_p=m$ and charge $q_p=e$.
2. An α-particle (helium nucleus) has a mass approximately 4 times that of a proton, so $m_{\alpha }=4m$, and a charge 2 times that of a proton, so $q_{\alpha }=2e$.

Let the accelerating potential for the proton be $V_p=V$ and the required accelerating potential for the α-particle be $V_{\alpha }$.

According to the problem, we want both particles to have the same de-Broglie wavelength:
${\lambda }_p={\lambda }_{\alpha }$

Substituting the values into the wavelength formula:
$\frac{h}{\sqrt{2m_p{q}_p{V}_p}}=\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }{V}_{\alpha }}}$

Squaring both sides and simplifying, we get:
$m_p{q}_p{V}_p=m_{\alpha }{q}_{\alpha }{V}_{\alpha }$

Now, substitute the relationships for mass and charge:
$m·e·V=\left(4m\right)·\left(2e\right)·V_{\alpha }$

Divide both sides by $m·e$:
$V=8V_{\alpha }$

Solving for $V_{\alpha }$ gives:
$V_{\alpha }=\frac{V}{8}$

Therefore, the required accelerating potential for the α-particle is V/8.