Question

The wavelength of the radiation emitted, when a hydrogen atom electron falls from infinity to stationary state 1, would be: (Rydberg constant \( R = 1.097 \times 10^7 \, \text{m}^{-1} \))

• A. \( 406 \, \text{nm} \)
• B. \( 192 \, \text{nm} \)
• C. \( 91 \, \text{nm} \)
• D. \( 9.1 \times 10^{-8} \, \text{nm} \)

Hint:

For transitions to the ground state (\( n_1 = 1 \)) from higher energy levels (\( n_2 = \infty \)), the wavelength is directly given by \( \lambda = \frac{1}{R} \), where \( R \) is the Rydberg constant. This gives the Lyman series wavelength, commonly 91 nm for the transition to \( n = 1 \).

Answer 1

The Correct Option is C

The wavelength \( \lambda \) of radiation can be calculated using the Rydberg formula for transitions from a higher energy level (\( n_2 \)) to the ground state (\( n_1 \)): \[ \frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \] where: - \( R = 1.097 \times 10^7 \, \text{m}^{-1} \) is the Rydberg constant, - \( n_1 = 1 \) (the final state, ground state), - \( n_2 = \infty \) (the initial state, which means the electron is at the ionization limit). ### Step 1: Apply the Rydberg formula For the transition from \( n_2 = \infty \) to \( n_1 = 1 \): \[ \frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{\infty^2}\right). \] Since \( \frac{1}{\infty^2} \) is zero, this simplifies to: \[ \frac{1}{\lambda} = R. \] ### Step 2: Substitute the value of \( R \) Substitute the value of the Rydberg constant: \[ \frac{1}{\lambda} = 1.097 \times 10^7 \, \text{m}^{-1}. \] ### Step 3: Solve for \( \lambda \) Now solve for \( \lambda \): \[ \lambda = \frac{1}{1.097 \times 10^7} \approx 91 \, \text{nm}. \] Final Answer: \[ \boxed{91 \, \text{nm}}. \]