JEE MAIN - Physics (2017 (Offline) - No. 24)
Some energy levels of a molecule are shown in the figure. The
ratio of the wavelengths r = $${{\lambda _1}}$$/$${{\lambda _2}}$$, is given by:
_en_24_1.png)
r = 1/3
r = 4/3
r = 2/3
r = 3/4
Explanation
From energy level diagram, using $$\Delta E = {{hc} \over \lambda }$$
For wavelength $${{\lambda _1}}$$, $$\Delta E =$$ –E – (–2E) = $${{hc} \over {{\lambda _1}}}$$
For wavelength $${{\lambda _2}}$$, $$\Delta E =$$ –E – (–$${{4E} \over 3}$$) = $${{hc} \over {{\lambda _2}}}$$
$$ \therefore $$ r = $${{{\lambda _1}} \over {{\lambda _2}}} = {{\left( {{{4E} \over 3} - E} \right)} \over {\left( {2E - E} \right)}}$$ = $${1 \over 3}$$
For wavelength $${{\lambda _1}}$$, $$\Delta E =$$ –E – (–2E) = $${{hc} \over {{\lambda _1}}}$$
For wavelength $${{\lambda _2}}$$, $$\Delta E =$$ –E – (–$${{4E} \over 3}$$) = $${{hc} \over {{\lambda _2}}}$$
$$ \therefore $$ r = $${{{\lambda _1}} \over {{\lambda _2}}} = {{\left( {{{4E} \over 3} - E} \right)} \over {\left( {2E - E} \right)}}$$ = $${1 \over 3}$$
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