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

The kinetic energy of an electron and a proton is $10^{-32} \ J$. Then the relation between their de-Broglie wavelengths is

Calculate the $(a)$ momentum,and $(b)$ de Broglie wavelength of the electrons accelerated through a potential difference of $56 \; V$.

An electron,accelerated by a potential difference $V$,has de Broglie wavelength $\lambda$. If the electron is accelerated by a potential difference $4V$,its de Broglie wavelength will become $\frac{\lambda}{n}$ where $n=$

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

An electron is accelerated from rest through a potential difference such that its kinetic energy is $1.5 \ eV$. The de-Broglie wavelength associated with this electron is: