Question

What is the shortest possible wavelength in the Lyman series in hydrogen?

Short answer

Answer: The shortest possible wavelength in the Lyman series of hydrogen is approximately 91.1 nm.

Step-by-step solution

Define the variables

We define the following variables to use in the Rydberg formula: - \(n_1 = 1\) (since the electron falls to the n=1 energy level for the Lyman series) - \(n_2 = \infty\) (as we are considering the shortest possible wavelength, which occurs when the electron falls from the highest possible energy level)

Plug the variables into the Rydberg formula

Substitute the values of \(n_1\) and \(n_2\) into the Rydberg formula: \( \frac{1}{\lambda} = R_H(\frac{1}{1^2} - \frac{1}{\infty^2}) \)

Simplify the equation

Simplify the equation by evaluating the terms within the parentheses: \( \frac{1}{\lambda} = R_H(1 - 0) \)

Solve for the wavelength

Solve for \(\lambda\) by multiplying both sides of the equation by \(R_H^{-1}\): \( \lambda = \frac{1}{R_H} \)

Calculate the shortest wavelength

Plug in the value of \(R_H\) and calculate the shortest wavelength: \( \lambda = \frac{1}{(1.097 \times 10^7 m^{-1})} \) \(\lambda \approx 9.11 \times 10^{-8} m\)

Convert the wavelength to nanometers

To express the result in nanometers, multiply the wavelength in meters by \(10^9\) : \( \lambda \approx (9.11 \times 10^{-8} m) \times (10^9 nm/m) \approx 91.1 nm \) The shortest possible wavelength in the Lyman series of hydrogen is approximately 91.1 nm.