Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ What would be the wavelength of the 2nd line in the Lyman series for the hydrogen atom, given [tex]R = 1.097 \times 10^7 \, \text{cm}^{-1}[/tex]?

A) 121.6 nm
B) 102.5 nm
C) 97.3 nm
D) 82.8 nm

The question is asking to calculate the wavelength of the 2nd line in the Lyman series for a hydrogen atom. The Rydberg formula is used to find the wavelength of lines in the emission spectrum of hydrogen:

1/λ = R × (1/n1² - 1/n2²), where λ is the wavelength, R is the Rydberg constant, n1 is the lower energy level, and n2 is the higher energy level. In the Lyman series, n1 is always 1, as electrons are falling back to the ground state. So for the 2nd line of Lyman series (Lyman beta), the electron falls from n2 = 3 to n1 = 1.

Using the given Rydberg constant value R = 1.097 × 10⁷ cm⁻¹ and the formula:

1/λ = (1.097 × 10⁷) × (1/1² - 1/3²)

1/λ = (1.097 × 10⁷) × (1 - 1/9)

1/λ = (1.097 × 10⁷) × (8/9)

λ = 9 / (8 × 1.097 × 10⁷) cm = 1.02 × 10⁻² cm = 102.5 nm

Therefore, the wavelength of the 2nd line in the Lyman series for the hydrogen atom is 102.5 nm.