The de-Broglie's wavelength of an electron present in the first Bohr orbit of an $H$ atom is:

$A$ stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are the charge and mass of an electron, respectively, then the value of $h/\lambda$ (where $\lambda$ is the wavelength associated with the electron wave) is given by:

Wavelength of $H^{+}$ ion with kinetic energy $1.65 \ eV$ is (mass of proton $= 1.6726 \times 10^{-27} \ kg$) (in $nm$)

The de-Broglie wavelength associated with a particle of mass $10^{-6} \ kg$ with a velocity of $10 \ ms^{-1}$ is $:-$

The wavelength (in $m$) of a particle of mass $1.67 \times 10^{-27} \ kg$ moving with velocity of $3.97 \times 10^6 \ m \ s^{-1}$ is:

The $K.E.$ of an electron is $4.55 \times 10^{-25} \,J$. Calculate its $\lambda$.