In an emission spectrum,the transitions n_2 t | vedclass

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(B) The energy of a photon emitted during an electronic transition in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 (\frac{1}{n_{lower}^2} - \

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$E_4 < E_3 < E_2 < E_1$

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(B) The energy of a photon emitted during an electronic transition in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 (\frac{1}{n_{lower}^2} - \frac{1}{n_{higher}^2}) ext{ eV}$.As the principal quantum number $n$ increases,the energy gap between consecutive shells decreases.Therefore,the energy difference for the transitions are:$E_1 (n=2 o n=1) > E_2 (n=3 o n=2) > E_3 (n=4 o n=3) > E_4 (n=5 o n=4)$.Thus,the increasing order of energies is $E_4 < E_3 < E_2 < E_1$.

In an emission spectrum,the transitions $n_2 \to n_1$,$n_3 \to n_2$,$n_4 \to n_3$,and $n_5 \to n_4$ emit energies $E_1, E_2, E_3$,and $E_4$ respectively. What is the increasing order of these energies?

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