One of the emission lines of the hydrogen atom has a wavelength of 94.974 nm. (b) Determine the initial and final values of n associated with this emission.

Verified step by step guidance

Identify the given wavelength of the emission line, which is 94.974 nm. Convert this wavelength into meters by multiplying by $10^{-9}$, as 1 nm = $10^{-9}$ meters.

Use the Rydberg formula for hydrogen emission, which is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, where $\lambda$ is the wavelength, $R_H$ is the Rydberg constant (approximately $1.097 \times 10^7 \, \text{m}^{-1}$), $n_1$ is the lower energy level, and $n_2$ is the higher energy level.

Calculate $\frac{1}{\lambda}$ using the converted wavelength in meters to find the inverse wavelength in meters$^{-1}$.

Rearrange the Rydberg formula to solve for $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ by dividing the inverse wavelength by the Rydberg constant.

Estimate possible values of $n_1$ and $n_2$ that satisfy the equation $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. Typically, $n_1$ will be a smaller integer (like 1 or 2) and $n_2$ will be a larger integer. Check different combinations of $n_1$ and $n_2$ to see which pair fits the calculated value from the previous step.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hydrogen Emission Spectrum

The hydrogen emission spectrum is a series of discrete wavelengths emitted by hydrogen atoms when electrons transition between energy levels. Each line in the spectrum corresponds to a specific transition, with the wavelength related to the energy difference between the initial and final states of the electron.

Energy Level Transitions

In an atom, electrons occupy quantized energy levels, denoted by the principal quantum number n. When an electron moves from a higher energy level (n_initial) to a lower one (n_final), it emits a photon with energy equal to the difference between these levels, which can be calculated using the Rydberg formula.

Rydberg Formula

The Rydberg formula allows for the calculation of the wavelengths of spectral lines in hydrogen. It is expressed as 1/λ = R_H (1/n_final² - 1/n_initial²), where R_H is the Rydberg constant. This formula is essential for determining the initial and final quantum numbers associated with a given wavelength.

Recommended video:

Guided course

02:26

Skeletal Formula

Related Practice

Textbook Question

The Lyman series of emission lines of the hydrogen atom are those for which nf = 1. (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed.

Textbook Question

The Lyman series of emission lines of the hydrogen atom are those for which nf = 1. (b) Calculate the wavelengths of the first three lines in the Lyman series—those for which ni = 2, 3, and 4.

Textbook Question

One of the emission lines of the hydrogen atom has a wavelength of 94.974 nm. (a) In what region of the electromagnetic spectrum is this emission found?

views

Textbook Question

The hydrogen atom can absorb light of wavelength 1094 nm. (a) In what region of the electromagnetic spectrum is this absorption found?

Textbook Question

The hydrogen atom can absorb light of wavelength 1094 nm. (b) Determine the final value of n associated with this absorption.

Textbook Question

Order the following transitions in the hydrogen atom from smallest to largest frequency of light absorbed: n = 3 to n = 7, n = 4 to n = 8, n = 2 to n = 5, and n = 1 to n = 3.

views