Example 3 Find the wavelength emitted by a helium ion He^+ during the transition of electron from the second energy level to the first (um)

To determine the wavelength emitted by a helium ion ($He^{+}$) during the transition of an electron from the second energy level to the first, we can use the Rydberg formula for hydrogen-like atoms:<br /><br />\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]<br /><br />where:<br />- \(\lambda\) is the wavelength of the emitted photon,<br />- \(R\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)),<br />- \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and higher energy levels, respectively.<br /><br />For the transition from the second energy level (\(n_2 = 2\)) to the first energy level (\(n_1 = 1\)) in a helium ion (\(He^{+}\)), the formula becomes:<br /><br />\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]<br /><br />\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) \]<br /><br />\[ \frac{1}{\lambda} = R \left( \frac{3}{4} \right) \]<br /><br />\[ \lambda = \frac{4}{3R} \]<br /><br />Now, substituting the value of the Rydberg constant:<br /><br />\[ \lambda = \frac{4}{3 \times 1.097 \times 10^7 \, \text{m}^{-1}} \]<br /><br />\[ \lambda \approx 1.022 \times 10^{-7} \, \text{m} \]<br /><br />\[ \lambda \approx 102.2 \, \text{nm} \]<br /><br />Therefore, the wavelength emitted by a helium ion ($He^{+}$) during the transition of an electron from the second energy level to the first is approximately 102.2 nanometers (nm).