Which of the following relations is correct, if the wavelength $(\lambda)$ is equal to the distance travelled by the electron in one second? $h$ is the Planck's constant and $m$ is the mass of electron.

$A$ 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 $Li^{3+} = 8.3 \times \text{mass of proton}$)

If the total energy of an electron in a hydrogen atom in an excited state is $-3.4 \text{ eV}$,then the de-Broglie wavelength of the electron is

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

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

The de Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as,[$v_0$ is threshold frequency]: