The de-Broglie wavelength of an electron is the same as that of a $50 \ keV$ $X$-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \ MeV$ ).

The ratio of the de-Broglie wavelength of an $\alpha$-particle to that of a proton,when both are subjected to the same magnetic field such that the radii of their paths are equal,assuming the magnetic field induction vector $\vec{B}$ is perpendicular to the velocity vectors of the $\alpha$-particle and the proton,is:

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 $V$)?

If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor

An electron of mass $m$ has de-Broglie wavelength $\lambda$ when accelerated through potential difference $V$. When a proton of mass $M$ is accelerated through a potential difference of $9V$,the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage).

$A$ particle is dropped from a height $H$. The de Broglie wavelength of the particle depends on height as