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

$A$ 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 two particles $A$ and $B$ have de Broglie wavelengths $1 \ nm$ and $5 \ nm$ respectively. If the mass of $A$ is four times the mass of $B$,the ratio of kinetic energies of $A$ and $B$ would be

$A$ metal is irradiated with light of wavelength $660 \, nm$. Given that the work function of the metal is $1.0 \, eV$,the de Broglie wavelength of the ejected electron is close to

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

If $a_0$ is the Bohr radius,then the de-Broglie wavelength of an electron revolving in the second excited state of the $H$ atom will be: