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:

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 :

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)

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