The mass of a photon with a wavelength equal to $1.54 \times 10^{-8} \ cm$ is

$A$ 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 wavelength of a microscopic particle of mass $9.1 \times 10^{-31} \ kg$ is $182 \ nm$. Its kinetic energy in $J$ is: $(h = 6.625 \times 10^{-34} \ J \ s)$

What will be the wavelength of a ball of mass $0.1 \,kg$ moving with a velocity of $10 \,m \,s^{-1}?$

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

When an electron in a hydrogen atom jumps from the third excited state to the ground state,the de-Broglie wavelength associated with the electron becomes