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

If the de Broglie wavelength of a dust particle of mass $1.0 \times 10^{-9} \,kg$ is $3 \times 10^{-25} \,m$, then the speed of the particle is . . . . . . . $\left(h=6.625 \times 10^{-34} \,J \,s\right)$

Explain the particle-wave (dual) nature of matter.

$A$ proton and an electron have the same de Broglie wavelength. If $K_p$ and $K_e$ are the kinetic energies of the proton and electron respectively,then choose the correct relation:

The kinetic energy of a free electron increases to $3$ times the previous kinetic energy $(K.E.)$. The ratio of the new de-Broglie wavelength to the previous de-Broglie wavelength is:

The de-Broglie wavelength of an electron is the same as that of a $50 \ keV$ $X$-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \ MeV$ ).