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(D) The energy of the emitted photon is equal to the difference in energy levels.For the transition from $2E$ to $E$:$\Delta E_1 = 2E - E = E = \frac{

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$3\lambda$

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(D) The energy of the emitted photon is equal to the difference in energy levels.For the transition from $2E$ to $E$:$\Delta E_1 = 2E - E = E = \frac{hc}{\lambda} \Rightarrow E = \frac{hc}{\lambda} \quad ... (1)$For the transition from $\frac{4E}{3}$ to $E$:$\Delta E_2 = \frac{4E}{3} - E = \frac{E}{3} = \frac{hc}{\lambda'} \quad ... (2)$Substituting the value of $E$ from equation $(1)$ into equation $(2)$:$\frac{1}{3} \left( \frac{hc}{\lambda} \right) = \frac{hc}{\lambda'}$$\frac{1}{3\lambda} = \frac{1}{\lambda'} \Rightarrow \lambda' = 3\lambda$Therefore, the wavelength of the photon produced is $3\lambda$.

The following diagram indicates the energy levels of a certain atom. When the system moves from the $2E$ level to the $E$ level, it emits a photon of wavelength $\lambda$. What is the wavelength of the photon produced during its transition from the $\frac{4E}{3}$ level to the $E$ level?

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