Problem

The series limit wavelength of the Balmer series is emitted as the electron in the hydrogen atom falls from $n = \infty$ to $n = 2$. What is the wavelength of this line if $\Delta E = 3.040 \, \mathrm{eV}$?

Data

• Energy Difference $\Delta E$: $3.040 \, \mathrm{eV}$
• Energy Level $n$: $\infty$
• Energy Level $p$: $2$
• Rydberg Constant $R_H$: $1.0974 \times 10^{7} \, \mathrm{m}^{-1}$

To find:

• Wavelength $\lambda$

Prerequisite Concepts

1. Balmer Series Formula:
   The wavelength of emitted photons in the Balmer 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. Series Limit Wavelength:
   The shortest wavelength in the Balmer series corresponds to $n = \infty$, where $\frac{1}{n^2} \to 0$.

3. Units Conversion:
   Convert the result to nanometers ($\mathrm{nm}$) for standard 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 $R_H = 1.0974 \times 10^{7} \, \mathrm{m}^{-1}$, $p = 2$, and $n = \infty$:

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

Step 3: Simplify the expression

Since $\frac{1}{\infty^2} = 0$:

$$\frac{1}{\lambda} = 1.0974 \times 10^{7} \left( \frac{1}{4} - 0 \right)$$ $$\frac{1}{\lambda} = 1.0974 \times 10^{7} \cdot 0.25$$ $$\frac{1}{\lambda} = 0.27435 \times 10^{7} \, \mathrm{m}^{-1}$$

Step 4: Calculate $\lambda$

$$\lambda = \frac{1}{0.27435 \times 10^{7}} \, \mathrm{m}$$ $$\lambda = 3.644979 \times 10^{-7} \, \mathrm{m}$$

Step 5: Convert to nanometers

$$\lambda = 364.49 \, \mathrm{nm}$$

Answer

The wavelength of the Balmer series limit line is:

$$\lambda = 364.49 \, \mathrm{nm}$$