In an emission spectrum,the transitions n_2 t | vedclass

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(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = R_H 	imes hc 	imes (\frac{1}{n_1^2} - \frac{1}{n_f^2})$.S

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$E_1 < E_2 < E_3 < E_4$

Vedclass · Answer

(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = R_H 	imes hc 	imes (\frac{1}{n_1^2} - \frac{1}{n_f^2})$.Since $n_1$ is constant for all transitions,the energy $\Delta E$ increases as the value of the initial energy level $n_f$ increases.For the given transitions:$E_1$ corresponds to $n=2 	o n=1$$E_2$ corresponds to $n=3 	o n=1$$E_3$ corresponds to $n=4 	o n=1$$E_4$ corresponds to $n=5 	o n=1$As $n$ increases from $2$ to $5$,the energy difference increases.Therefore,the increasing order of energies is $E_1 < E_2 < E_3 < E_4$.

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

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