If the kinetic energy of a particle is reduced to half,the de-Broglie wavelength becomes

The de-Broglie wavelength $(\lambda)$ for electron $(e)$, proton $(p)$ and $He^{2+}$ ion $(\alpha)$ are in the following order. (Speed of $e$, $p$ and $\alpha$ are the same)

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

Calculate the mass of a photon with wavelength $3.6 \, \mathring{A}$.

Number of waves formed by a Bohr electron in one complete revolution in its third orbit is