For the Balmer series in the spectrum of H at | vedclass

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(B) For the Balmer series,$n_{1}=2$ and $n_{2}=3, 4, 5, \dots, \infty$.$(I)$ As wavelength $\lambda$ decreases,the wave number $\bar{v}$ increases,and

Vedclass · Accepted Answer

$(I)$,$(II)$,$(III)$

Vedclass · Answer

(B) For the Balmer series,$n_{1}=2$ and $n_{2}=3, 4, 5, \dots, \infty$.$(I)$ As wavelength $\lambda$ decreases,the wave number $\bar{v}$ increases,and the lines converge towards the series limit,so this is correct.$(II)$ By definition of the Balmer series,$n_{1}=2$,so this is correct.$(III)$ The wavelength $\lambda$ is longest when the energy difference $\Delta E$ is minimum,which occurs for the transition $n_{2}=3$ to $n_{1}=2$. Thus,this is correct.$(IV)$ The ionization energy corresponds to the transition from $n=1$ to $n=\infty$. While the Balmer series lines provide $R_{H}$,the ionization energy is specifically calculated using the Rydberg constant $R_{H}$ and the ground state energy,not directly from the wave numbers of the Balmer lines alone. Therefore,this statement is generally considered incorrect in this context.Thus,statements $(I)$,$(II)$,and $(III)$ are correct.

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