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

The radiation corresponding to $$3 \to 2$$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field $$3 \times {10^{ - 4}}\,T.$$ If the radius of the larger circular path followed by these electrons is $$10.0$$ $$mm$$, the work function of the metal is close to:

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?

In a hydrogen like atom electron make transition from an energy level with quantum number $$n$$ to another with quantum number $$\left( {n - 1} \right)$$. If $$n > > 1,$$ the frequency of radiation emitted is proportional to :