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

Two particles of masses $m$ and $2m$ have equal kinetic energies. The de-Broglie wavelengths are in the ratio of

The velocity associated with a proton moving in a potential difference of $1000 \, V$ is $4.37 \times 10^{5} \, ms^{-1}$. If a hockey ball of mass $0.1 \, kg$ is moving with this velocity,calculate the wavelength associated with this velocity.

The de Broglie wavelength of a particle of mass $1 \ mg$ moving with a velocity of $10 \ m \ s^{-1}$ is $(h=6.63 \times 10^{-34} \ J \ s)$

The velocities of two particles $A$ and $B$ are $0.05 \ ms^{-1}$ and $0.02 \ ms^{-1}$ respectively. The mass of $B$ is five times the mass of $A$. The ratio of their de-Broglie's wavelength is

If the velocity of the electron in Bohr's first orbit is $2.19 \times 10^{6} \, m s^{-1}$, calculate the de Broglie wavelength associated with it.