Calculate the wavelength (in nanometer) associated with a proton moving at $1.0 \times 10^3 \ m \ s^{-1}$. (Mass of proton $= 1.67 \times 10^{-27} \ kg$ and $h = 6.63 \times 10^{-34} \ J \ s$)

Calculate the momentum of a particle which has a wavelength of $1 \ \mathring{A}$.

$A$ body of mass $x$ $kg$ is moving with a velocity of $100$ $ms^{-1}$. Its de-Broglie wavelength is $6.62 \times 10^{-35}$ $m$. Hence,$x$ is ($h = 6.62 \times 10^{-34}$ $Js$). (in $kg$)

$A$ $0.66 \ kg$ ball is moving with a speed of $100 \ m/s$. The associated wavelength will be $(h = 6.6 \times 10^{-34} \ J \ s)$.

The energy of separation of an electron in a $H$ atom in an excited state is $3.4 \ eV$. The de-Broglie wavelength (in $\mathring{A}$) associated with the above electron is,if the radius of the first orbit of the $H$ atom is $0.53 \ \mathring{A}$.

$A$ cricket ball of $0.5 \ kg$ is moving with a velocity of $100 \ m/s$. The wavelength associated with its motion is