An $\alpha$ particle and a proton are accelerated from rest by a potential difference of $200 \ V$. After this,their de Broglie wavelengths are $\lambda_{\alpha}$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_{\alpha}}$ is

$A$ 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:

How has the scanning tunneling microscope become useful in estimating the size of an atom?

An electron of mass $m$ when accelerated through a potential difference $V$ has a de-Broglie wavelength $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be:

An electron,a doubly ionized helium ion $(He^{++})$ and a proton have the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_{e}, \lambda_{He^{++}}$ and $\lambda_{P}$ is:

An electron (mass $m$) with initial velocity $\overrightarrow{v} = v_{0} \hat{i} + v_{0} \hat{j}$ is in an electric field $\overrightarrow{E} = -E_{0} \hat{k}$. If $\lambda_{0}$ is the initial de-Broglie wavelength of the electron,its de-Broglie wavelength at time $t$ is given by: