A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :

Half-lives of two radioactive elements $$A$$ and $$B$$ are $$20$$ minutes and $$40$$ minutes, respectively. Initially, the samples have equal number of nuclei. After $$80$$ minutes, the radio of decayed number of $$A$$ and $$B$$ nuclei will be:

As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion :

Hydrogen $$\left( {{}_1{H^1}} \right)$$, Deuterium $$\left( {{}_1{H^2}} \right)$$, singly ionised Helium $${\left( {{}_2H{e^4}} \right)^ + }$$ and doubly ionised lithium $${\left( {{}_3L{i^6}} \right)^{ + + }}$$ all have one electron around the nucleus. Consider an electron transition from $$n=2$$ to $$n=1.$$ If the wavelengths of emitted radiation are $${\lambda _1},{\lambda _2},{\lambda _3}$$ and $${\lambda _4}$$ respectively then approximately which one of the following is correct?