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(B) The energy of an emitted photon is given by $\Delta E = \frac{hc}{\lambda}$, where $h$ is Planck's constant and $c$ is the speed of light.For the

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$\frac{4}{9}$

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(B) The energy of an emitted photon is given by $\Delta E = \frac{hc}{\lambda}$, where $h$ is Planck's constant and $c$ is the speed of light.For the transition from $4E$ to $E$:$\Delta E_1 = 4E - E = 3E$Therefore, $3E = \frac{hc}{\lambda_1} \implies \lambda_1 = \frac{hc}{3E}$ ... $(i)$For the transition from $\frac{7}{3}E$ to $E$:$\Delta E_2 = \frac{7}{3}E - E = \frac{4}{3}E$Therefore, $\frac{4}{3}E = \frac{hc}{\lambda_2} \implies \lambda_2 = \frac{3hc}{4E}$ ... $(ii)$Taking the ratio of $\lambda_1$ to $\lambda_2$:$\frac{\lambda_1}{\lambda_2} = \frac{hc / 3E}{3hc / 4E} = \frac{hc}{3E} 	imes \frac{4E}{3hc} = \frac{4}{9}$Thus, the ratio $\frac{\lambda_1}{\lambda_2}$ is $\frac{4}{9}$.

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