JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 2)
Both the nucleus and the atom of some element arein their respective first excited states. They get de-excted by emitting photons of wavelengths $$\lambda $$N, $$\lambda $$A respectively. The ratio $${{{}^\lambda N} \over {{}^\lambda A}}$$is closest to :
10$$-$$6
10
10$$-$$10
10$$-$$1
Explicació
We know that $$E = {{hc} \over \lambda }$$
So, for atom $${E_A} = {{hc} \over {{\lambda _A}}}$$
And for neutron $${E_N} = {{hc} \over {{\lambda _N}}}$$
Then, $${{{E_A}} \over {{E_N}}} = {{hc} \over {{\lambda _A}}} \times {{{\lambda _N}} \over {hc}} \Rightarrow {{{\lambda _N}} \over {{\lambda _A}}}$$
Here, EA is order of eV and EN is order of MeV.
Therefore, $${{{\lambda _N}} \over {{\lambda _A}}} = {{{E_A}} \over {{E_N}}} = {{eV} \over {MeV}} = {{1eV} \over {{{10}^6}eV}}$$
$${\lambda _N} = {10^{ - 6}}\,{\lambda _A}$$