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 :

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 $$ = 9 \times {10^{ - 31}}\,kg$$ ).

A diatomic molecule is made of two masses $${m_1}$$ and $${m_2}$$ which are separated by a distance $$r.$$ If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by: ($$n$$ is an integer)