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(B) The energy of a photon emitted during a transition between two energy levels $E_i$ and $E_f$ is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda

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$\frac{4}{9}$

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(B) The energy of a photon emitted during a transition between two energy levels $E_i$ and $E_f$ is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda}$.For the transition from $4E$ to $E$:$\Delta E_1 = 4E - E = 3E$So,$3E = \frac{hc}{\lambda_1} \implies \lambda_1 = \frac{hc}{3E}$  --- $(1)$For the transition from $\frac{7}{3}E$ to $E$:$\Delta E_2 = \frac{7}{3}E - E = \frac{4}{3}E$So,$\frac{4}{3}E = \frac{hc}{\lambda_2} \implies \lambda_2 = \frac{3hc}{4E}$  --- $(2)$Taking the ratio of $\lambda_1$ to $\lambda_2$:$\frac{\lambda_1}{\lambda_2} = \left( \frac{hc}{3E} ight) / \left( \frac{3hc}{4E} ight) = \frac{hc}{3E} 	imes \frac{4E}{3hc} = \frac{4}{9}$

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