Question

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition? a. \(n=4 \rightarrow n=3\) b. \(n=5 \rightarrow n=4\) c. \(n=5 \rightarrow n=3\)

Short answer

The wavelengths of light emitted during the transitions in a hydrogen atom are as follows: \(a. \lambda_a \approx 1.88 \times 10^{-6} \: \text{m}\), \(b. \lambda_b \approx 4.32 \times 10^{-7} \: \text{m}\), and \(c. \lambda_c \approx 1.28 \times 10^{-6} \: \text{m}\). The corresponding types of electromagnetic radiation are: a. Infrared (IR) b. Visible c. Infrared (IR)

Step-by-step solution

Understanding the Rydberg formula

The Rydberg formula is used to calculate the wavelength of light emitted/absorbed during electron transitions in an atom. It is given by: \[\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\] Where: - \(\lambda\) is the wavelength of the light - \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final states, respectively - \(R_H\) is the Rydberg constant for hydrogen, approximately equal to \(1.097 \times 10^7 \: \text{m}^{-1}\)

Calculate the wavelength for each transition

We will now use the Rydberg formula to find the wavelength for each given transition. a. For the transition \(n=4 \rightarrow n=3\): \[\frac{1}{\lambda_a} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right)\] \[\lambda_a = \frac{1}{R_H \left( \frac{1}{9} - \frac{1}{16} \right)}\] b. For the transition \(n=5 \rightarrow n=4\): \[\frac{1}{\lambda_b} = R_H \left( \frac{1}{4^2} - \frac{1}{5^2} \right)\] \[\lambda_b = \frac{1}{R_H \left( \frac{1}{16} - \frac{1}{25} \right)}\] c. For the transition \(n=5 \rightarrow n=3\): \[\frac{1}{\lambda_c} = R_H \left( \frac{1}{3^2} - \frac{1}{5^2} \right)\] \[\lambda_c = \frac{1}{R_H \left( \frac{1}{9} - \frac{1}{25} \right)}\]

Finding the type of electromagnetic radiation

Now that we have the wavelengths, we can identify the corresponding type of electromagnetic radiation: a. For the transition \(n=4 \rightarrow n=3\): Calculate \(\lambda_a\) using the Rydberg constant: \[\lambda_a = \frac{1}{1.097 \times 10^7 \: \text{m}^{-1} \left( \frac{1}{9} - \frac{1}{16} \right)}\] \[\lambda_a \approx 1.88 \times 10^{-6} \: \text{m}\] Since \(\lambda_a\) falls within the range of \(7 \times 10^{-7} \: \text{m}\) to \(10^{-3} \: \text{m}\), the radiation is Infrared (IR). b. For the transition \(n=5 \rightarrow n=4\): Calculate \(\lambda_b\) using the Rydberg constant: \[\lambda_b = \frac{1}{1.097 \times 10^7 \: \text{m}^{-1} \left( \frac{1}{16} - \frac{1}{25} \right)}\] \[\lambda_b \approx 4.32 \times 10^{-7} \: \text{m}\] Since \(\lambda_b\) falls within the range of \(4 \times 10^{-7} \: \text{m}\) to \(7 \times 10^{-7} \: \text{m}\), the radiation is Visible. c. For the transition \(n=5 \rightarrow n=3\): Calculate \(\lambda_c\) using the Rydberg constant: \[\lambda_c = \frac{1}{1.097 \times 10^7 \: \text{m}^{-1} \left( \frac{1}{9} - \frac{1}{25} \right)}\] \[\lambda_c \approx 1.28 \times 10^{-6} \: \text{m}\] Since \(\lambda_c\) falls within the range of \(7 \times 10^{-7} \: \text{m}\) to \(10^{-3} \: \text{m}\), the radiation is Infrared (IR). In conclusion, the types of electromagnetic radiation emitted for each transition are as follows: a. Infrared (IR) b. Visible c. Infrared (IR)