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(D) The wavelength $\lambda$ of the spectral line emitted during a transition between energy levels is given by the Rydberg formula: $\frac{1}{\lambda

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Di-ionized lithium

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(D) The wavelength $\lambda$ of the spectral line emitted during a transition between energy levels is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For a transition from $n_2 = 2$ to $n_1 = 1$,the term $\left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4}$ is constant for all hydrogen-like species.Therefore,$\frac{1}{\lambda} \propto Z^2$,which implies $\lambda \propto \frac{1}{Z^2}$.To obtain the minimum wavelength $\lambda$,the atomic number $Z$ must be maximum.- Hydrogen atom $(H)$: $Z = 1$- Deuterium atom $(D)$: $Z = 1$- Uni-ionized helium $(He^+)$: $Z = 2$- Di-ionized lithium $(Li^{2+})$: $Z = 3$Since $Li^{2+}$ has the highest atomic number $(Z = 3)$,it produces the spectral line with the minimum wavelength.

In the following atoms and molecules,for the transition from $n = 2$ to $n = 1$,the spectral line of minimum wavelength will be produced by:

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