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

Two base balls (masses: $m_1 = 100 \ g$ and $m_2 = 50 \ g$) are thrown. Both of them move with uniform velocity, but the velocity of $m_2$ is $1.5$ times that of $m_1$. The ratio of de Broglie wavelengths $\lambda(m_1) : \lambda(m_2)$ is given by

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

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

$A$ 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 wavelength of an electron and a neutron will become equal when the velocity of the electron is $x$ times the velocity of the neutron. The value of $x$ is (Nearest Integer). (Mass of electron is $9.1 \times 10^{-31} \ kg$ and mass of neutron is $1.6 \times 10^{-27} \ kg$)