Using the Rydberg formula, calculate the wavelength of the second line in the Balmer series of the hydrogen…

Question

Using the Rydberg formula, calculate the wavelength of the second line in the Balmer series of the hydrogen spectrum.

ScanSolve AI Tutor · Accepted Answer

To find the wavelength of the second line in the Balmer series of the hydrogen spectrum, we use the Rydberg formula:

\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

where:

$\lambda$ is the wavelength
$R$ is the Rydberg constant for hydrogen ( $1.097 \times 10^{-2} \text{ nm}^{-1}$ )
$n_f$ is the final principal quantum number
$n_i$ is the initial principal quantum number

Identify the given values.
For the Balmer series, the electron transitions to the $n_f = 2$ energy level.
The "second line" in the Balmer series corresponds to a transition from $n_i = 4$ to $n_f = 2$ . (The first line is $n_i=3 \to n_f=2$ , the second is $n_i=4 \to n_f=2$ , etc.)
The Rydberg constant $R = 1.097 \times 10^{-2} \text{ nm}^{-1}$ .
Substitute the values into the Rydberg formula.
$\frac{1}{\lambda} = (1.097 \times 10^{-2} \text{ nm}^{-1}) \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$
Calculate the terms inside the parenthesis.
$\frac{1}{\lambda} = (1.097 \times 10^{-2} \text{ nm}^{-1}) \left( \frac{1}{4} - \frac{1}{16} \right)$ $\frac{1}{\lambda} = (1.097 \times 10^{-2} \text{ nm}^{-1}) \left( \frac{4}{16} - \frac{1}{16} \right)$ $\frac{1}{\lambda} = (1.097 \times 10^{-2} \text{ nm}^{-1}) \left( \frac{3}{16} \right)$ $\frac{1}{\lambda} = (1.097 \times 10^{-2} \text{ nm}^{-1}) (0.1875)$
Multiply by the Rydberg constant.
$\frac{1}{\lambda} = 0.002056875 \text{ nm}^{-1}$
Calculate the wavelength $\lambda$ .
$\lambda = \frac{1}{0.002056875 \text{ nm}^{-1}}$ $\lambda \approx 486.19 \text{ nm}$
Comparing this value to the given options, $486.19 \text{ nm}$ is closest to $486.3 \text{ nm}$ .

The final answer is $\text{c. 486.3nm}$ .

Using the Rydberg formula, calculate the wavelength of the second line in the Balmer series of the hydrogen spectrum.