Answer :
Final answer:
The wavelength of the second line of the Lyman series for a hydrogen atom is approximately 101.6 nm, which rounds to option b) 102.8 nm when expressing the answer with the options given.
Explanation:
The wavelength of the second line of the Lyman series in the spectrum of a hydrogen atom can be calculated using the Rydberg formula, which is defined as:
\(\frac{1}{\lambda} = R \cdot (\frac{1}{n_1^2} - \frac{1}{n_2^2})\)
For the Lyman series, \(n_1 = 1\). The second line signifies a transition from \(n_2 = 3\) to \(n_1 = 1\). Using the Rydberg constant \(R = 1.107 \times 10^7 m^{-1}\), we can calculate:
\(\frac{1}{\lambda} = 1.107 \times 10^7 \cdot (1 - \frac{1}{3^2})\)
\(\frac{1}{\lambda} = 1.107 \times 10^7 \cdot (1 - \frac{1}{9})\)
\(\frac{1}{\lambda} = 1.107 \times 10^7 \cdot (\frac{8}{9})\)
\(\frac{1}{\lambda} = 9.839 \times 10^6 m^{-1}\)
\(\lambda = \frac{1}{9.839 \times 10^6} \approx 1.016 \times 10^{-7} m\)
Converting to nanometers (1 m = 10^9 nm), we get:
\(\lambda \approx 101.6 nm\)
Therefore, the correct answer is b) 102.8 nm, as it is the closest to our calculated value.
