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

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

If the de Broglie wavelength of an electron is $2 \ nm$,then its kinetic energy is nearly (Planck's constant $= 6.6 \times 10^{-34} \ J \ s$ and mass of electron $= 9 \times 10^{-31} \ kg$) (in $eV$)

An electron (mass $m$) with an initial velocity $v = v_0 \hat{i}$ is in an electric field $E = E_0 \hat{j}$. If $\lambda_0 = h/mv_0$,its de Broglie wavelength at time $t$ is given by

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

Calculate the de Broglie wavelength of an electron in the first Bohr orbit of a hydrogen atom if the velocity of an electron in the first orbit is $2.2 \times 10^6 \text{ m s}^{-1}$. [mass of electron = $9.1 \times 10^{-31} \text{ kg}$, Planck's constant $(h)$ = $6.626 \times 10^{-34} \text{ J s}$]