Question 211: Dual Nature of Matter and Radiation

Question

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 :

Accepted Answer

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, $${{{E_A}} \over {{E_N}}} = {{hc} \over {{\lambda _A}}} 	imes {{{\lambda _N}} \over {hc}} \Rightarrow {{{\lambda _N}} \over {{\lambda _A}}}$$
Here, E
A
is order of eV and E
N
is order of MeV.
Therefore, $${{{\lambda _N}} \over {{\lambda _A}}} = {{{E_A}} \over {{E_N}}} = {{eV} \over {MeV}} = {{1eV} \over {{{10}^6}eV}}$$
$${\lambda _N} = {10^{ - 6}}\,{\lambda _A}$$

Answer

10$$-$$6

Answer

10

Answer

10$$-$$10

Answer

10$$-$$1

Dual Nature of Matter and Radiation

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 :

Choose the correct answer: