The electronic transitions from n=2 to n=1 wi | vedclass

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(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = Z^2 \cdot R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^

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$Li^{2+}$

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(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = Z^2 \cdot R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$ For the transition from $n=2$ to $n=1$: $\frac{1}{\lambda} = Z^2 \cdot R_H \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = Z^2 \cdot R_H \left( \frac{3}{4} \right)$ This implies $\lambda = \frac{4}{3 Z^2 R_H}$, so $\lambda \propto \frac{1}{Z^2}$. To obtain the shortest wavelength, the atomic number $Z$ must be the maximum. Comparing the species: $H$ $(Z=1)$, $He^{+}$ $(Z=2)$, and $Li^{2+}$ $(Z=3)$. Since $Li^{2+}$ has the highest $Z$, it will produce the shortest wavelength.

The electronic transitions from $n=2$ to $n=1$ will produce the shortest wavelength in which of the following species? (where $n$ is the principal quantum state)

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