The wavelength associated with an electron accelerated through a potential difference of $100 \ V$ is nearly .............. $ \mathring A $

Hydrogen and helium atoms are at temperatures of $27^{\circ}C$ and $127^{\circ}C$ respectively. Find the ratio of their de Broglie wavelengths.

Compute the typical de Broglie wavelength of an electron in a metal at $27\,^{\circ} C$ and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{-10} \; m$.

An electron microscope uses electrons of $40 \ keV$. The de Broglie wavelength associated with these electrons is approximately:

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to

The de Broglie wavelength of a particle accelerated through a potential difference of $150 \ V$ is $10^{-10} \ m$. If it is accelerated through a potential difference of $600 \ V$,what will be its wavelength in $\mathring A$?