A proton and an alpha particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is
(A) $\sqrt 8 $.
(B) $\dfrac{1}{{\sqrt 8 }}$.
(C) $1$.
(D) $2$.

Hint: The de Broglie wavelength of a particle denotes the length of the scale at which a wave-like properties are important for that particle. De Broglie wavelength is usually represented by the symbol $\lambda $.

Useful formula
The de Broglie equation for wavelength is given by;
$\lambda = \dfrac{h}{{\sqrt {2mK} }}$
Where, $\lambda $ denotes the de-Broglie wavelength of the particle, $h$ denotes the planck's constant, $m$ denotes the mass of the particle.

Complete step by step solution
According to de Broglie equation for wavelength is given by;
$
  \lambda = \dfrac{h}{{\sqrt {2mK} }} \\
  \lambda = \dfrac{h}{{\sqrt {2mqV} }} \\
 $
We already know that $K = qV$
Since the value for $V$ is the same for both alpha and the proton particle
That is the ratio of de-Broglie wavelengths of the proton and the alpha particle is calculated as;
$\dfrac{{{\lambda _\alpha }}}{{{\lambda _p}}} = \sqrt {\dfrac{{{m_\alpha }{q_\alpha }}}{{{m_p}{q_p}}}} $
Where, ${\lambda _\alpha }$ denotes the wavelength of the alpha particle, ${\lambda _p}$ denotes the wavelength of the proton particle, ${m_\alpha }$ denotes the mass of the alpha particle, ${m_p}$ denotes the mass of the proton particle.
$
  \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \sqrt {\dfrac{{4{m_p} \times 2{q_p}}}{{{m_p}{q_p}}}} \\
  \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \sqrt 8 \\
 $

Therefore, the ratio of de-Broglie wavelengths of the proton and the alpha particle is given as $\sqrt 8 $.

Hence, the option (A) $\sqrt 8 $ is the correct answer.

Note de Broglie equation explains that a matter can act as waves as much as light and radiation which also perform as waves and particles. The equation further explains that a beam of electrons can also be diffracted just like a beam of light.