The ratio of wavelengths for the transition of electrons from the $2^{nd}$ orbit to the $1^{st}$ orbit of Helium $(He^+)$ and Lithium $(Li^{++})$ is (Atomic number of Helium = $2$,Atomic number of Lithium = $3$).

The de Broglie wavelength of the electron in the ground state is $\lambda_1$ and that in the $n = 3$ level is $\lambda_3$. Then $\lambda_3$ is given by:

$A$ photon of wavelength $300 \ nm$ interacts with a stationary hydrogen atom in the ground state. During the interaction, the whole energy of the photon is transferred to the electron of the atom. State which possibility is correct. (Consider, Planck constant $= 4 \times 10^{-15} \ eV \cdot s$, velocity of light $= 3 \times 10^{8} \ m/s$, ionisation energy of hydrogen $= 13.6 \ eV$)

Magnetic field at the centre of a hydrogen-like atom (atomic number $= Z$) due to the motion of an electron in the $n^{th}$ orbit is proportional to

The ionization energy of a $Li^{++}$ ion is .......

The radius of the first orbit of hydrogen is $r_{H}$,and the energy in the ground state is $-13.6 \text{ eV}$. Considering a $\mu^{-}$-particle with a mass $207 m_e$ revolving around a proton as in a hydrogen atom,the energy and radius of the proton and $\mu^{-}$-combination respectively in the first orbit are (assume the nucleus to be stationary):