A proton and an $\alpha - particle$are accelerated through the same potential difference.Which one of the two has
(i) greater de-Broglie wavelength, and (ii) less kinetic energy? Justify your answer.

Hint: Charge and mass of both proton and $\alpha - particle$ are different so they will have different electrostatic potential energy or kinetic energy and since energies and masses are different there will be difference in de-broglie wavelength too.

Complete answer:
We know that when a charge particle q is accelerated through a potential difference V then the charge particle acquires a kinetic energy $qV$. So, the charge on the proton is $e$. And charge on $\alpha - particle$ is 2e. Both are accelerated through same potential difference (let's say it is V)

Then the kinetic energy of the proton will be $eV$. And the kinetic energy of $\alpha - particle$ will be $2eV$. As we can clearly see proton has less kinetic energy.Now we will calculate de-broglie wavelength.Let mass of proton is $m$, then mass of $\alpha - particle$ will be $4m$.

We know that for an elementary particle de-broglie wavelength.
$\lambda = \dfrac{h}{{\sqrt {2mKE} }}$
Where $m$ is mass of the particle and $KE$ is kinetic energy of the particle
So,
${\lambda _{proton}} = \dfrac{h}{{\sqrt {2meV} }}$ and
${\lambda _{\alpha - particle}} = \dfrac{h}{{\sqrt {2(4m)2eV} }}\\
\therefore{\lambda _{\alpha - particle}} = \dfrac{h}{{\sqrt {16meV} }}$
Clearly proton has greater de-broglie wavelength.

Hence the correct answer for (i) is proton and for (ii) is also proton.

Note: It looks counter-intuitive that answer for both parts is same which is proton in this case but it should be true because charge on proton is less than the $\alpha - particle$ so kinetic energy of the proton must be smaller as compared to $\alpha - particle$ and since kinetic energy is smaller de-broglie wavelength must be larger for proton.