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

What is the de-Broglie wavelength of the $\alpha$-particle accelerated through a potential difference $V$?

$(a)$ For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40 \times 10^{-10} \; m ?$
$(b)$ Also find the de Broglie wavelength of a neutron,in thermal equilibrium with matter,having an average kinetic energy of $\frac{3}{2} k T$ at $300 \; K$.

If $E_p$ and $E_e$ represent the kinetic energy of a photon and an electron respectively. If the de-Broglie wavelength $\lambda_p$ of a photon is twice the de-Broglie wavelength $\lambda_e$ of an electron,then $E_e / E_p$ is (Speed of electron $= C/100$,where $C$ is the velocity of light).

If the momentum of an electron is changed by $P_m$ and the associated de Broglie wavelength changes by $0.50\ \%$, find the initial momentum of the electron. (in $P_m$)

An electron accelerated through a potential of $10000 \ V$ from rest has a de-Broglie wavelength $\lambda$. What should be the accelerating potential,so that the wavelength is doubled (in $V$)?