$A$ proton and an electron are accelerated by the same potential difference. If their de-Broglie wavelengths are $\lambda_p$ and $\lambda_e$ respectively,then:

The figure shows four situations in which an electron is moving in an electric or magnetic field. In which case is the de Broglie wavelength of the electron increasing?

The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is

The de-Broglie wavelength of an electron moving with a velocity $v = c / 2$ ($c$ is the velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of the electron and the photon is:

$A$ 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

The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of $3 \ km/s$ will be ............ $\mathring{A}$.