In the hydrogen atom, the electron makes a transition from the higher orbit $(i)$ to a lower orbit $(f)$. The ratio of the radius of the orbits is given by $r_i : r_f = 16 : 4$. The wavelength of the photon emitted due to this transition is . . . . . . nm. (Given Rydberg constant $R = 1.0973 \times 10^7 \text{ m}^{-1}$)

$A$ 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 $.80$)

Let $A_n$ be the area enclosed by the $n^{th}$ orbit in a hydrogen atom. The graph of $\ln (A_n/A_1)$ against $\ln (n)$:

In a muonic atom,a muon of mass $200$ times that of an electron and the same charge is bound to a proton. The wavelengths of its Balmer series are in the range of

In Bohr's model of the hydrogen atom,let $PE$ represent potential energy and $TE$ represent total energy. When an electron transitions to a higher energy level,

The ratio of areas within the electron orbits for the first excited state to the ground state for a hydrogen atom is (in $ : 1$)