Which of the transitions in a hydrogen atom e | vedclass

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(B) The frequency of the emitted photon is given by $
u = \frac{\Delta E}{h}$,where $\Delta E$ is the energy difference between the two levels. Since

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$n = 4$ to $n = 3$

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(B) The frequency of the emitted photon is given by $
u = \frac{\Delta E}{h}$,where $\Delta E$ is the energy difference between the two levels. Since $
u \propto \Delta E$,the lowest frequency corresponds to the transition with the smallest energy difference.From the energy level diagram:For $n = 2$ to $n = 1$: $\Delta E = -3.4 - (-13.6) = 10.2 \, eV$For $n = 4$ to $n = 3$: $\Delta E = -0.85 - (-1.51) = 0.66 \, eV$For $n = 3$ to $n = 1$: $\Delta E = -1.51 - (-13.6) = 12.09 \, eV$For $n = 4$ to $n = 2$: $\Delta E = -0.85 - (-3.4) = 2.55 \, eV$The transition $n = 4$ to $n = 3$ has the smallest energy difference $(0.66 \, eV)$,and therefore emits a photon of the lowest frequency. Thus,the correct option is $B$.

Which of the transitions in a hydrogen atom emits a photon of the lowest frequency? ($n =$ quantum number)

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