Problem

What is the wavelength of the second line of the Paschen series?

Data

• Energy level $n$: $5$
• Energy level $p$: $3$
• Rydberg Constant $R_H$: $1.0974 \times 10^{7} \, \mathrm{m}^{-1}$

To find:

• Wavelength ($\lambda$)

Prerequisite Concepts

1. Paschen Series Formula:
   The wavelength of emitted photons in the Paschen series is given by:

$$\frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$$

Where:

• $R_H$: Rydberg constant
• $p$: Lower energy level
• $n$: Higher energy level ($n > p$)

2. Units:
   The result is typically converted into nanometers (nm) for easier interpretation.

Solution

Step 1: Write the general formula

$$\frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$$

Step 2: Substitute the given values

Substitute $R_H = 1.0974 \times 10^{7} \, \mathrm{m}^{-1}$, $p = 3$, and $n = 5$:

$$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{1}{3^2} - \frac{1}{5^2} \right)$$

Step 3: Simplify the expression

$$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{1}{9} - \frac{1}{25} \right)$$

Find the difference:

$$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{25 - 9}{225} \right)$$ $$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{16}{225} \right)$$

Simplify further:

$$\frac{1}{\lambda} = 0.718 \times 10^{7} \, \mathrm{m}^{-1}$$

Step 4: Calculate $\lambda$

$$\lambda = \frac{1}{0.718 \times 10^{7}} \, \mathrm{m}$$ $$\lambda = 1.281 \times 10^{-6} \, \mathrm{m}$$

Step 5: Convert to nanometers

$$\lambda = 1281.4 \, \mathrm{nm}$$

Answer

The wavelength of the second line of the Paschen series is:

$$\lambda = 1281.4 \, \mathrm{nm}$$