Problem

Calculate the longest wavelength of radiation for the Paschen series.

Data

• Energy Level $n$: $4$
• Energy Level $p$: $3$
• Rydberg’s 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 calculated using:

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

Where:

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

2. Longest Wavelength:
   Occurs when the difference between the energy levels is smallest, i.e., $n = 4$ and $p = 3$.

3. Units Conversion:
   The result is typically converted into nanometers ($\mathrm{nm}$) for clarity.

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 $n = 4$:

$$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$$

Step 3: Simplify the expression

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

Find the difference:

$$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{16 - 9}{144} \right)$$ $$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{7}{144} \right)$$ $$\frac{1}{\lambda} = 0.5334 \times 10^{7} \, \mathrm{m}^{-1}$$

Step 4: Calculate $\lambda$

$$\lambda = \frac{1}{0.5334 \times 10^{7}} \, \mathrm{m}$$ $$\lambda = 1.8747655 \times 10^{-6} \, \mathrm{m}$$

Step 5: Convert to nanometers

$$\lambda = 1874.8 \, \mathrm{nm}$$

Answer

The longest wavelength of radiation for the Paschen series is:

$$\lambda = 1874.8 \, \mathrm{nm}$$