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 :

Hydrogen atom is excited from ground state to another state with principal quantum number equal to $$4.$$ Then the number of spectral lines in the emission spectra will be :

Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (mass of neutron $$ = 1.6725 \times {10^{ - 27}}kg,$$ mass of proton $$ = 1.6725 \times {10^{ - 27}}\,kg,$$ mass of electron $$ = 9 \times {10^{ - 31}}\,kg$$ ).