The energy level diagram for a hydrogen-like atom is shown in the figure. The radius of its first Bohr orbit is

Potential energy between a proton and an electron is given by $U = \frac{K e^2}{3 R^3}$. Then,the radius of the Bohr orbit can be given by:

The magnetic field at the centre due to the motion of an electron in the first Bohr orbit is $B$. The magnetic field due to the motion of an electron in the second Bohr orbit at the centre will be

What is ground state?

If the radius of the first Bohr orbit is $r$, then the radius of the second Bohr orbit will be

$A$ hydrogen atom,initially at rest in its ground state,absorbs a photon of frequency $v_1$ and ejects the electron with a kinetic energy of $10 eV$. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency $v_2$. The center of mass of the resulting positronium atom moves with a kinetic energy of $5 eV$. It is given that the positron has the same mass as that of the electron and the positronium atom can be considered as a Bohr atom,in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process,the difference between the two photon energies (in $eV$) is $....$ (in $.80$)