Final answer:
To calculate the wavelength of the spectral line produced in a hydrogen atom when an electron transitions from n=3 to n=2, we use the Rydberg formula and find that the wavelength is 656 nm, which corresponds to red light in the visible spectrum.
Step-by-step explanation:
To calculate the wavelength of the spectral line produced when an electron in a hydrogen atom transitions from energy level n=3 to n=2, we can use the Rydberg formula for the hydrogen emission spectrum:
\(\frac{1}{\lambda} = R_\infty \left(\frac{1}{n^2_1} - \frac{1}{n^2_2}\right)\)
where:
- \(\lambda\) is the wavelength of the emitted light,
- \(R_\infty\) is the Rydberg constant (1.097373 × 10^7 m^-1),
- \(n_1\) and \(n_2\) are the initial and final energy levels of the electron (2 and 3 in this case).
Inserting the energy levels into the equation:
\(\frac{1}{\lambda} = 1.097373 × 10^7 m^-1 \left(\frac{1}{2^2} - \frac{1}{3^2}\right)\)
Calculating the difference:
\(\frac{1}{\lambda} = 1.097373 × 10^7 m^-1 \left(\frac{1}{4} - \frac{1}{9}\right)\)
\(\frac{1}{\lambda} = 1.097373 × 10^7 m^-1 \cdot \frac{5}{36}\)
Now we calculate \(\lambda\):
\(\lambda = \frac{1}{1.097373 × 10^7 m^-1 \cdot \frac{5}{36}}\)
\(\lambda = 6.56 × 10^-7 m\) or 656 nm
This wavelength is part of the visible light spectrum, specifically it is red light, which corresponds to the transition given by the Balmer series for hydrogen.