Which of the following graphs correctly represents the de-Broglie wavelength $\lambda$ associated with a fundamental particle having linear momentum $p$?

$A$ proton and an $\alpha$-particle are accelerated using the same potential difference. How are the de-Broglie wavelengths $\lambda_p$ and $\lambda_{\alpha}$ related to each other?

$A$ particle of mass $2 \times 10^{-27} \,kg$ has a de-Broglie wavelength of $3.3 \times 10^{-10} \,m$. The kinetic energy of this particle is (Planck's constant $h = 6.6 \times 10^{-34} \,J \cdot s$).

The potential difference $V$ required for accelerating an electron to have the de-Broglie wavelength of $1 \text{ Å}$ is (in $\text{ V}$)

$A$ proton and an $\alpha$-particle are accelerated from rest to the same kinetic energy. The ratio of their de Broglie wavelengths $\lambda_{p} : \lambda_{\alpha}$ is:

The graph between the energy log $E$ of an electron and its de$-$Broglie wavelength log $\lambda$ will be