The de-Broglie equation is:

If the de Broglie wavelength of an electron is $728.14 \ nm$,its kinetic energy in $J$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

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:

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 $0.53 \ \mathring{A}$.

What is the de Broglie wavelength of a baseball with a mass of $120 \ g$ moving at a velocity of $44.7 \ m \ s^{-1}$?

With what velocity must an electron travel,so that its momentum is equal to that of a photon of wavelength $663 \ nm$ (in $m/s$)?