To calculate the wavelength of light for the electronic transition in a hydrogen atom from $n = 3$ to $n = 5$, we use the Rydberg formula:

$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Where:

• $\lambda$ is the wavelength,
• $R_H$ is the Rydberg constant, $R_H = 1.097 \times 10^7 \, \text{m}^{-1}$,
• $n_1$ is the lower energy level (3 in this case),
• $n_2$ is the higher energy level (5 in this case).

Step-by-step calculation:

1. Substitute the values: $\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{5^2} \right)$ $\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{25} \right)$

2. Simplify the fractions: $\frac{1}{9} = 0.1111, \quad \frac{1}{25} = 0.04$ $\frac{1}{9} - \frac{1}{25} = 0.1111 - 0.04 = 0.0711$

3. Now calculate $\frac{1}{\lambda}$: $\frac{1}{\lambda} = 1.097 \times 10^7 \times 0.0711 = 7.801 \times 10^5 \, \text{m}^{-1}$

4. To find $\lambda$, take the reciprocal: $\lambda = \frac{1}{7.801 \times 10^5} \, \text{m} = 1.282 \times 10^{-6} \, \text{m}$

5. Convert the wavelength to nanometers (1 m = $10^9$ nm): $\lambda = 1.282 \times 10^{-6} \, \text{m} \times 10^9 \, \text{nm/m} = 1282 \, \text{nm}$

Thus, the wavelength of the electronic transition from $n = 3$ to $n = 5$ in the hydrogen atom is 1282 nm.

Would you like more details or have any questions?

Related Questions:

1. How does the Rydberg formula apply to other atoms beyond hydrogen?
2. What is the significance of energy level transitions in hydrogen spectroscopy?
3. How does the wavelength change if the transition occurs between other levels, such as $n = 2$ and $n = 4$?
4. What are the visible wavelengths for transitions in the hydrogen atom?
5. Can we calculate the energy corresponding to this wavelength?

Tip:

Remember that transitions between higher energy levels (like $n = 3$ to $n = 5$) typically result in wavelengths in the infrared region, while transitions involving lower levels might be in the visible or ultraviolet range.