$A$ proton is accelerated through a potential difference of $100 \ V$. To have the same de Broglie wavelength,what potential difference must be applied across a doubly ionized $_4^8 Be$ nucleus (in $V$)?

$A$ proton moving with one-tenth of the velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having a certain kinetic energy has the same de Broglie wavelength $\lambda$. The ratio of the kinetic energy of the proton to that of the alpha particle is:

The energy that should be added to an electron,to reduce its de-Broglie wavelength from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$,will be

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$,their de Broglie wavelength in the frame of reference attached to their centre of mass is

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

The ratio of wavelengths of a proton and a deuteron accelerated by potentials $V_{p}$ and $V_{d}$ is $1 : \sqrt{2}$. Then,the ratio of $V_{p}$ to $V_{d}$ will be: