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

If $a_0$ is the radius of the first Bohr's orbit of the $H$-atom, the de-Broglie wavelength of an electron revolving in the second Bohr's orbit will be: (in $\pi a_0$)

If the kinetic energy of an electron is increased four times,the wavelength of the de-Broglie wave associated with it would become

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 $\lambda$ of an electron in the $2^{nd}$ Bohr orbit is:

If the velocities of hydrogen $(H_2)$ and helium $(He)$ molecules are equal,then the ratio of the wavelengths of helium to hydrogen is .......