$A$ proton and an $\alpha$-particle are accelerated through the same potential difference $V$. The ratio of their de-Broglie wavelengths is

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 energy that should be added to an electron to reduce its de Broglie wavelength from $1 \, nm$ to $0.5 \, nm$ is

The de Broglie wavelength of a proton accelerated by a potential difference of $100 \ V$ is $\lambda_0$. If an alpha particle is accelerated by the same potential difference,its de Broglie wavelength will be:

The de-Broglie wavelength of an electron with kinetic energy of $320 eV$ is (Take $h = 6.0 \times 10^{-34} \text{ SI unit}$, mass of electron $m_{e} = 9.0 \times 10^{-31} \text{ kg}$, charge of an electron $e = 1.6 \times 10^{-19} \text{ C}$). (in $pm$)

What is the $(a)$ momentum,$(b)$ speed,and $(c)$ de Broglie wavelength of an electron with kinetic energy of $120 \ eV$?