The de-Broglie wavelength of a body of mass $1 \ kg$ moving with a velocity of $2000 \ m/s$ is:

For the same objective,find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000\,\mathring{A}$ and electrons accelerated through $100\,V$ used as the illuminating substance.

The de-Broglie wavelength of an electron having $80 eV$ energy is nearly ($1 eV = 1.6 \times 10^{-19} J$,Mass of the electron $= 9 \times 10^{-31} kg$,Planck's constant $= 6.6 \times 10^{-34} J-s$). (in $Å$)

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

The wavelength of a very fast-moving electron $(v \approx c)$ is:

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 $n=$