Question

Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths $r={\lambda }_1/{\lambda }_2$, is given by

• A. $r=\frac{1}{3}$
• B. $r=\frac{4}{3}$
• C. $r=\frac{2}{3}$
• D. $r=\frac{3}{4}$

Answer 1

The correct option is A $r=\frac{1}{3}$

$E=\frac{hc}{\lambda }$
$\Rightarrow \lambda =\frac{hc}{E}$
$\Rightarrow \lambda \propto \frac{1}{E}$
From figure,
Energy of photon of wavelength ${\lambda }_2$ $△E_2=-E-(-\frac{4}{3}E)$
$=\frac{E}{3}$
Energy of photon of wavelength ${\lambda }_1$ $△E_1=-E-(-2E)$
$=E$
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{△E_2}{△E_1}=\frac{\left(\frac{E}{3}\right)}{E}$
$r=\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{3}$