The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of $3 \ km/s$ will be ............ $\mathring{A}$.

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

The de Broglie wavelength of an electron moving with kinetic energy of $144 \;eV$ is nearly

$A$ moving proton and electron have the same de Broglie wavelength. If ${K}$ and ${P}$ denote the kinetic energy and momentum respectively,then choose the correct option:

The de Broglie wavelength and kinetic energy of a particle are $2000 \ \mathring{A}$ and $1 \ \text{eV}$ respectively. If its kinetic energy becomes $1 \ \text{MeV}$,then its de Broglie wavelength becomes $...... \ \mathring{A}$.

An electron beam has a kinetic energy equal to $100\; eV$. Find its wavelength (in $\mathring{A}$) associated with the beam,given the mass of an electron $= 9.1 \times 10^{-31}\; kg$,$1\; eV = 1.6 \times 10^{-19}\; J$,and Planck's constant $= 6.6 \times 10^{-34}\; Js$.