To calculate the wavelength of light emitted during an electron transition in a hydrogen atom, we use the Rydberg formula:

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

Where:

• $\lambda$ is the wavelength of the emitted light,
• $R_H$ is the Rydberg constant, $R_H = 1.097 \times 10^7 \, \text{m}^{-1}$,
• $n_i$ is the initial energy level (in this case $n_i = 8$),
• $n_f$ is the final energy level (in this case $n_f = 5$).

Step 1: Calculate the wavelength

We can substitute the values into the equation:

$\frac{1}{\lambda} = 1.097 \times 10^7 \, \text{m}^{-1} \left( \frac{1}{5^2} - \frac{1}{8^2} \right)$

$\frac{1}{\lambda} = 1.097 \times 10^7 \, \text{m}^{-1} \left( \frac{1}{25} - \frac{1}{64} \right)$

Step 2: Simplify the expression

First, find a common denominator:

$\frac{1}{25} - \frac{1}{64} = \frac{64 - 25}{1600} = \frac{39}{1600}$

So, the equation becomes:

$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{39}{1600}$

$\frac{1}{\lambda} = 2.673 \times 10^5 \, \text{m}^{-1}$

Step 3: Calculate the wavelength

$\lambda = \frac{1}{2.673 \times 10^5} \, \text{m}$

$\lambda = 3.74 \times 10^{-6} \, \text{m}$

Step 4: Convert to nanometers

Since $1 \, \text{m} = 10^9 \, \text{nm}$:

$\lambda = 3.74 \times 10^{-6} \, \text{m} \times 10^9 \, \text{nm/m} = 3740 \, \text{nm}$

Thus, the wavelength is 3740 nm, which corresponds to option A.

Answer: A. 3740 nm

Would you like further clarification or details?

---

Here are 5 related questions:

1. How is the energy of a photon related to its wavelength?
2. What is the significance of the Rydberg constant in quantum physics?
3. How does the Bohr model of the hydrogen atom explain spectral lines?
4. How do principal quantum numbers affect energy levels in an atom?
5. What is the Balmer series, and how does it relate to visible hydrogen emissions?

Tip: When dealing with light transitions, higher energy differences result in shorter wavelengths, which corresponds to ultraviolet or visible light.