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Some energy levels of a molecule are shown in the figure with their wavelengths of transitions. Then:
A
$\lambda_3>\lambda_2, \lambda_1=2\lambda_2$
B
$\lambda_3>\lambda_2, \lambda_1=4\lambda_2$
C
$\lambda_1>\lambda_2, \lambda_2=2\lambda_3$
D
$\lambda_2>\lambda_1, \lambda_2=2\lambda_3$
Detailed Solution
From the energy level diagram: $\dfrac{hc}{\lambda_1}=-\dfrac{5E}{2}+4E=\dfrac{3E}{2}$; $\dfrac{hc}{\lambda_2}=-2E+3E=E$; $\dfrac{hc}{\lambda_3}=-2E+4E=2E$. Comparing $\lambda_2,\lambda_3$: $\dfrac{1}{\lambda_3}=\dfrac{2}{\lambda_2}\Rightarrow\lambda_2=2\lambda_3$. Comparing $\lambda_1,\lambda_2$: $3\lambda_1=2\lambda_2\Rightarrow\lambda_1\lambda_1,\ \lambda_2=2\lambda_3$.