Find the de-Broglie wavelength of an electron with kinetic energy of $ 120 eV $. (in $pm$)

The graph shows the variation of de-Broglie wavelength $\lambda$ versus $\frac{1}{\sqrt{V}}$,where $V$ is the accelerating potential for four particles carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents the particle with the smallest mass?

The de-Broglie wavelength of an electron accelerated by a potential of $50\,V$ is approximately ............. $\mathring{A}$. $(|e| = 1.6 \times 10^{-19}\,C, m_e = 9.1 \times 10^{-31}\,kg, h = 6.6 \times 10^{-34}\,Js)$

An electron and a proton are accelerated through the same potential difference. The ratio of the de-Broglie wavelength $\lambda_{p}$ to $\lambda_{e}$ is $[m_{e} = \text{mass of electron}, m_{p} = \text{mass of proton}]$

$A$ particle is moving along the $x-$axis back and forth in a box of length $L$. Assuming the de-Broglie hypothesis is applicable for the particle,and it is moving with a constant speed in the box making perfectly elastic collisions with the walls. The possible value of the momentum of the particle is (Note: $h$ is Planck's constant,$n$ is the principal quantum number).

The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is