To calculate the wavelength of light emitted when an electron moves from $n = 5$ to $n = 4$ in a hydrogen atom, we can use the Rydberg formula for hydrogen:

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

where:

• $\lambda$ is the wavelength of the emitted light,
• $R_H$ is the Rydberg constant for hydrogen, $R_H = 1.097 \times 10^7 \, \text{m}^{-1}$,
• $n_1$ is the lower energy level (in this case $n_1 = 4$),
• $n_2$ is the higher energy level (in this case $n_2 = 5$).

Step-by-step solution:

1. Substitute the values into the Rydberg formula: $\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4^2} - \frac{1}{5^2} \right)$

2. Calculate the terms inside the parentheses: $\frac{1}{4^2} = \frac{1}{16} = 0.0625$ $\frac{1}{5^2} = \frac{1}{25} = 0.04$ $\frac{1}{16} - \frac{1}{25} = 0.0625 - 0.04 = 0.0225$

3. Now substitute this difference into the equation: $\frac{1}{\lambda} = 1.097 \times 10^7 \times 0.0225 = 2.46825 \times 10^5 \, \text{m}^{-1}$

4. Finally, take the reciprocal to find $\lambda$: $\lambda = \frac{1}{2.46825 \times 10^5} = 4.052 \times 10^{-6} \, \text{m} = 4052 \, \text{nm}$

Answer:

The wavelength of light produced is $\mathbf{4050 \, \text{nm}}$ (rounded to three significant figures).

Would you like more details or further clarifications?

Here are 5 related questions:

1. What is the Rydberg constant, and why is it important in hydrogen spectrum calculations?
2. How does the energy difference between electron levels relate to the wavelength of emitted light?
3. Why do different transitions in hydrogen produce different wavelengths?
4. How does the wavelength of light change with higher $n$-level transitions in hydrogen?
5. What other series of transitions exist in the hydrogen atom besides this one?

Tip: Transitions to the lowest energy levels (like $n=1$) produce light in the ultraviolet range, while transitions between higher levels (e.g., $n=5$ to $n=4$) produce infrared light.