The de Broglie wavelength associated with a ball of mass $1 \, kg$ having kinetic energy $0.5 \, J$ is

The ratio of de-Broglie wavelengths of two particles having mass ratio $1:3$ and kinetic energy ratio $2:1$ is

If the kinetic energy of an electron is $18.2 \times 10^{-25} \ J$,its de Broglie wavelength in $nm$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

$A$ 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 kinetic energy of an electron is $4.55 \times 10^{-25} \ J$ and its mass is $9.1 \times 10^{-31} \ kg$. Calculate the velocity,momentum,and wavelength of the electron.

The de Broglie wavelength of a car of mass $1000 \ kg$ and velocity $36 \ km/hr$ is