Problem

Find the shortest wavelength photon emitted in the Lyman series of hydrogen.

Data

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

To find:

• Wavelength ($\lambda$)

Prerequisite Concepts

1. Lyman Series Formula: The wavelength of emitted photons in the hydrogen spectrum 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. Shortest Wavelength: Occurs when $n \to \infty$, as $\frac{1}{n^2} \to 0$.

Solution

Step 1: Substitute the given values into the formula

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

Given:

• $p = 1$
• $n = \infty$

Substitute $n = \infty$:

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

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

$$\frac{1}{\lambda} = R_H \cdot 1$$ $$\frac{1}{\lambda} = R_H$$

Step 2: Calculate $\lambda$

Substitute $R_H = 1.0974 \times 10^{7} \, \mathrm{m}^{-1}$:

$$\lambda = \frac{1}{R_H}$$ $$\lambda = \frac{1}{1.0974 \times 10^{7}} \, \mathrm{m}$$ $$\lambda = 9.11 \times 10^{-8} \, \mathrm{m}$$

Step 3: Convert to nanometers (nm)

$$\lambda = 91.1 \, \mathrm{nm}$$

Answer

The shortest wavelength photon emitted in the Lyman series of hydrogen is:

$$\lambda = 91 \, \mathrm{nm}$$