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

An electron,accelerated by a potential difference $V$,has de Broglie wavelength $\lambda$. If the electron is accelerated by a potential difference $4V$,its de Broglie wavelength will become $\frac{\lambda}{n}$ where $n=$

$A$ charged particle is accelerated from rest through a certain potential difference. The de-Broglie wavelength is $\lambda_1$ when it is accelerated through $V_1$ and is $\lambda_2$ when accelerated through $V_2$. The ratio $\lambda_1 / \lambda_2$ is

Let $E_e$ and $E_p$ represent the kinetic energy of an electron and a photon,respectively. If the de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron,then find the ratio $E_p / E_e$. (Given: speed of electron $v = c / 100$,where $c$ is the velocity of light).

The wavelength of a charged particle of mass $8.0 \times 10^{-31} \ kg$, charge $1.6 \times 10^{-19} \ C$ and kinetic energy $3 \ keV$ will be (Planck constant, $h = 6.4 \times 10^{-34} \ Js$) (in $\text{Å}$)

What does the de Broglie equation for an electron represent?