Problem 57 - College Physics

Question

(a) What is the least amount of energy, in electron volts, that must be given to a hydrogen atom which is initially in its ground level so that it can emit the \(\mathrm{H}_{\alpha}\) line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the \(n=3\) level and eventually ends up in the ground level? Calculate the wavelength of the emitted photon in each case.

Step-by-Step Solution

Answer

(a) The energy required is 12.09 eV. (b) Two emissions: H-alpha at 656 nm, direct at 102.6 nm.

Step 1: Understanding the Balmer Series

The Balmer series in the hydrogen atom corresponds to electron transitions from higher energy levels down to the second energy level (\(n=2\)). The \(\mathrm{H}_{\alpha}\) line specifically corresponds to the transition from \(n=3\) to \(n=2\).

Step 2: Energy Requirement for Emitting \(\mathrm{H}_{\alpha}\)

To emit the \(\mathrm{H}_{\alpha}\) line, the hydrogen atom's electron must first be excited from the ground state \(n=1\) to at least the \(n=3\) state. The energy difference needed can be calculated using the Rydberg formula: \[E_n = -13.6 \left( \frac{1}{n^2} \right) \text{ eV}\]The energy required to excite from \(n=1\) to \(n=3\) is given by:\[E = E_1 - E_3 = 13.6 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 13.6 \times \frac{8}{9} = 12.09 \text{ eV}\].

Step 3: Analyzing Transitions from \(n=3\) to \(n=1\)

From the \(n=3\) level, the hydrogen atom can transition to \(n=2\) and then to \(n=1\), or directly to \(n=1\). Thus, there are two possible emission lines: \(n=3 \rightarrow n=2\) and \(n=3 \rightarrow n=1\).

Step 4: Calculating Wavelengths of Emitted Photons

Using the Rydberg formula for wavelength of emitted light:\[\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\]where \(R = 1.097 \times 10^7 \text{ m}^{-1}\):1. For the \(n=3 \rightarrow n=2\) transition:\[\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{2^2} - \frac{1}{3^2} \right) = 1.5243 \times 10^6 \text{ m}^{-1}\]\(\lambda \approx 656 \text{ nm}\) (\(\mathrm{H}_{\alpha}\) line).2. For the \(n=3 \rightarrow n=1\) transition:\[\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{1^2} - \frac{1}{3^2} \right) = 9.758 \times 10^6 \text{ m}^{-1}\]\(\lambda \approx 102.6 \text{ nm}\), which is in the ultraviolet range.

Key Concepts

Balmer Series

The Balmer Series is a set of spectral lines that arise when an electron in a hydrogen atom makes transitions from higher energy states to the second energy level, denoted as \( n=2 \). These spectral lines fall within the visible spectrum, making them particularly important in astronomy and spectroscopy.

Each line in the Balmer Series corresponds to a specific transition. The most notable among them is the \( \mathrm{H}_{\alpha} \) line, which results from an electron moving from the third energy level \( n=3 \) to \( n=2 \). This line produces a distinct red color and is commonly observed in emission spectra of hydrogen.

The other lines in the Balmer Series, such as \( \mathrm{H}_{\beta} \) and \( \mathrm{H}_{\gamma} \), involve transitions from higher levels like \( n=4 \) and \( n=5 \) respectively, down to \( n=2 \). The visible nature of these transitions makes the Balmer Series crucial in identifying hydrogen presence in astronomical bodies.

Rydberg Formula

The Rydberg Formula is a powerful tool for predicting the wavelengths of spectral lines in hydrogen and hydrogen-like ions. It relates the wavelength \( \lambda \) of the emitted or absorbed light to the energy levels involved in the transition.

The formula is given by:

• \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \)

where \( R_H \) 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.

This formula allows us to calculate the wavelength of light emitted during an electron transition. In the context of the Balmer Series, the Rydberg formula can determine each spectral line's exact wavelength, such as the \( \mathrm{H}_{\alpha} \) line's 656 nm wavelength.

Energy Levels

In an atom, electrons occupy specific regions of space called energy levels or shells. These levels are quantized, meaning electrons can only exist in certain, discrete energy states, each identified by the principal quantum number \( n \).

The energy \( E \) of these levels for hydrogen atoms is quantified by:

• \( E_n = -13.6 \left( \frac{1}{n^2} \right) \text{ eV} \)

where \( n \) is an integer starting from 1. The negative sign indicates that electrons are bound to the nucleus.

The ground state of a hydrogen atom corresponds to \( n=1 \), the lowest possible energy level. Excitation requires energy to be absorbed, elevating the electron to higher levels such as \( n=2 \) or \( n=3 \). Understanding these levels is crucial for calculating energy changes and emission or absorption spectra as electrons move between them.

Electron Transitions

Electron transitions occur when an electron absorbs or emits energy to move between different energy levels in an atom. These movements are what give rise to the spectra we observe.

For a transition to occur, an electron must gain or lose a precise amount of energy, equivalent to the energy difference between the initial and final levels. When an electron moves from a higher to a lower level, energy is released in the form of a photon, resulting in an emission line in the spectrum.

Considering a hydrogen atom starting at \( n=3 \), it might transition to \( n=2 \) (producing an \( \mathrm{H}_{\alpha} \) line) or directly to \( n=1 \), a move releasing more energy as ultraviolet light. Multiple transitions create different spectral lines, helping us understand the energy structure within the atom.

Calculating the energy and wavelength for these transitions involves understanding both energy levels and the Rydberg Formula, enabling predictions of the spectral characteristics observed scientifically.