One of the visible lines in the hydrogen emission spectrum corresponds to the \(n=6\) to \(n=2\) electronic transition. What color light is this transition? See Exercise 150 .

The Rydberg formula is used to calculate the wavelength of the light emitted due to electronic transitions in the hydrogen atom. The formula is given by: \( \dfrac{1}{\lambda} = R_{H} \left(\dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}}\right) \) Where: - \( \lambda \) is the wavelength of the light emitted, - \( R_{H} \) is the Rydberg constant for hydrogen, approximately equal to \( 1.097 \times 10^{7} \mathrm{m}^{-1} \), - \( n_{1} \) and \( n_{2} \) are the initial and final energy levels, respectively. In this case, we have the transitions from \( n_{1} = 6 \) to \( n_{2} = 2 \).

Using the Rydberg formula, we can calculate the wavelength of the light emitted during the transition from n=6 to n=2. \( \dfrac{1}{\lambda} = R_{H} \left(\dfrac{1}{2^{2}} - \dfrac{1}{6^{2}}\right) \) |Plug in the values and solve for \( \lambda \). \( \dfrac{1}{\lambda} = 1.097 \times 10^{7} \mathrm{m}^{-1} \left(\dfrac{1}{4} - \dfrac{1}{36}\right) \) \( \dfrac{1}{\lambda} = 1.097 \times 10^{7} \mathrm{m}^{-1} \left(\dfrac{32}{144}\right) \) \( \dfrac{1}{\lambda} = 1.097 \times 10^{7} \mathrm{m}^{-1} \times \dfrac{8}{36} \) Now, find \( \lambda \) by taking the inverse: \( \lambda = \dfrac{36}{1.097 \times 10^{7} \mathrm{m}^{-1} \times 8} \) \( \lambda \approx 4.103 \times 10^{-7} \mathrm{m} \) \( \lambda \approx 410.3 \mathrm{nm} \)

The color of the light can be determined by looking up the wavelength in a table of visible light colors or using an online tool. The wavelength of \( 410.3 \mathrm{nm} \) corresponds to the color violet, which lies around spectral range (400nm - 450nm). Therefore, the color of the light emitted during the transition from n=6 to n=2 is violet.