Question

The Pfund series of lines in the emission spectrum of hydrogen corresponds to transitions from higher excited states to the n = 5 orbit. Calculate the wavelength of the second line in the Pfund series to three significant figures. In which region of the spectrum does it lie?

Answer 1

Answer:

$\large \boxed{\text{4650 nm}}$

Explanation:

The second line in the Pfund series corresponds to a transition from n = 7 to n = 5.

To calculate the wavelength of the transition, we can use the Rydberg equation:

$\dfrac{1}{\lambda} = R_{H}\left ( \dfrac{1 }{n_{1}^{2}} - \dfrac{1 }{n_{2}^{2}} \right )$

where

$R_{H} = 1.097 \times 10^{7} \text{ m}^{-1}$

If n₁ = 5 and n₂ = 7

$\begin{array}{rcl}\dfrac{1}{\lambda} & = & 1.097 \times 10^{7} \text{ m}^{-1}\left ( \dfrac{1 }{5^{2}} - \dfrac{1 }{7^{2}} \right )\\\\ & = & 1.097 \times 10^{7} \text{ m}^{-1}\left ( \dfrac{1 }{25} - \dfrac{1 }{49} \right )\\\\ & = & 1.097 \times 10^{7} \text{ m}^{-1}\left ( \dfrac{49 - 25 }{49 \times25}\right )\\\\& = & 1.097 \times 10^{7} \text{ m}^{-1}\left ( \dfrac{24 }{1225}\right )\\\\ & = & 2.149 \times 10^{7}\text{ m}^{-1}\\\end{array}$

$\begin{array}{rcl}\lambda & = & \dfrac{1}{2.149 \times 10^{7}\text{ m}^{-1}}\\\\ & = & 4.65 \times 10^{-6} \text{ m}\\ & = & \mathbf{4650} \textbf{ nm}\\\end{array}\\\text{The wavelength of the line is \large \boxed{\textbf{4650 nm}} , which is in the \textbf{infrared region} }.$