The de Broglie wavelength of an electron travelling with $20 \%$ of velocity of light is
$(h = 6.626 \times 10^{-34} \ J \ s; m_{e} = 9.1 \times 10^{-31} \ kg)$

$A$ $20 \, g$ object is moving with a velocity of $100 \, ms^{-1}$. The de Broglie wavelength (in $m$) of the object is [Planck's constant $h = 6.626 \times 10^{-34} \, Js$]

$A$ proton and a $Li^{3+}$ nucleus are accelerated by the same potential. If $\lambda_{Li}$ and $\lambda_{P}$ denote the de Broglie wavelengths of $Li^{3+}$ and proton respectively,then the value of $\frac{\lambda_{Li}}{\lambda_{P}}$ is $x \times 10^{-1}$. The value of $x$ is ............
(Rounded off to the nearest integer)
(Mass of $Li^{3+} = 8.3 \times \text{mass of proton}$)

The de Broglie wavelength of a particle is $1000 \ nm$. What is its momentum? $(h = 6.6 \times 10^{-34} \ J \ s)$

The kinetic energy of a proton wave is $500 \ eV$. Calculate its de Broglie wavelength. (Mass of proton $= 1.67 \times 10^{-27} \ kg$)

The de-Broglie wavelength of a tennis ball of mass $60 \, g$ moving with a velocity of $10 \, m/s$ is approximately