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

To find 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:<br /><br />\[<br />\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)<br />\]<br /><br />where:<br />- $\lambda$ is the wavelength of the emitted light,<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 ($n_1 = 1$), we have:<br /><br />\[<br />\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R<br />\]<br /><br />Now, we can solve for $\lambda$:<br /><br />\[<br />\lambda = \frac{4}{3R}<br />\]<br /><br />Substituting the value of the Rydberg constant:<br /><br />\[<br />\lambda = \frac{4}{3 \times 1.097 \times 10^7 \, \text{m}^{-1}} \approx 2.92 \times 10^{-7} \, \text{m} = 292 \, \text{nm}<br />\]<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 292 nm.