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.

The velocity $v$ of a de Broglie wave is given by
$\left[ \begin{array}{c} u = \text{Frequency} \\ m = \text{mass} \\ C = \text{Velocity of light} \end{array} \right]$

If the radius of the first orbit of $H$ atom is $a_0$, what is the de-Broglie wavelength of an electron in the third orbit (in $\pi a_0$)?

$A$ 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 atomic masses of $He$ and $Ne$ are $4$ and $20 \ amu$,respectively. The value of the de Broglie wavelength of $He$ gas at $-73^{\circ} C$ is $M$ times that of the de Broglie wavelength of $Ne$ at $727^{\circ} C$. $M$ is

The de-Broglie wavelength associated with a particle of mass $10^{-6} \ kg$ with a velocity of $10 \ ms^{-1}$ is $:-$