The following diagram indicates the energy le | vedclass

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Question

$2 \mathrm{E}-\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \Rightarrow \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$

$\frac{4 \mathrm{E}}{3}-\mathrm{E}=\frac{\

Vedclass · Accepted Answer

$3\lambda $

Vedclass · Answer

$2 \mathrm{E}-\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \Rightarrow \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$

$\frac{4 \mathrm{E}}{3}-\mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}} \Rightarrow \frac{\mathrm{E}}{3}=\frac{\mathrm{hc}}{\lambda^{\prime}} $

$	herefore \frac{\lambda^{\prime}}{\lambda}=3 \Rightarrow \lambda^{\prime}=3 \lambda$

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

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