If ${\lambda _p}$ and ${\lambda _\alpha }$ are the de-Broglie wavelengths associated with protons and $\alpha$-particles of equal kinetic energies,then:

What is the de Broglie wavelength of an electron accelerated through a potential difference of $ 100 \ V $ (in $\text{Å}$)?

What is the experimental value of the de-Broglie wavelength for an electron accelerated through a potential difference of $V$ volts?

The wavelength $\lambda_e$ of an electron and $\lambda_p$ of a photon of same energy $E$ are related by

Particle $A$ of mass $m_{A} = \frac{m}{2}$ moving along the $x$-axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B} = \frac{m}{3}$. If both particles move along the $x$-axis after the collision,the change $\Delta \lambda$ in the de-Broglie wavelength of particle $A$,in terms of its de-Broglie wavelength $(\lambda_{0})$ before the collision is:

The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)