Problem 63 - University Physics with Modern Physics

Question

(a) What is the smallest amount of energy in electron volts that must be given to a hydrogen atom initially in its ground level so that it can emit the \(\mathrm{H}_{a}\) 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 emilted photon in each case.

Step-by-Step Solution

Answer

(a) 12.1 eV; (b) 3 transitions: 656.3 nm, 102.6 nm, and 121.5 nm.

Step 1: Understanding the Problem

To solve this problem, we need to determine (a) the minimum energy required to excite a hydrogen atom from its ground state to a level where it can emit an \(\text{H}_{\alpha}\) line and (b) the number of possible transitions and their emitted photon wavelengths when it returns to the ground state from \(n=3\).

Step 2: Identify Key Concepts

The \(\text{H}_{\alpha}\) line in the Balmer series corresponds to an electron transition from \(n=3\) to \(n=2\). To emit this line, the electron must first be excited to at least \(n=3\) from \(n=1\). The energy levels are given by \(E_n = -13.6 \frac{1}{n^2}\) eV.

Step 3: Calculate Energy for \(\text{H}_{\alpha}\) Emission

Calculate the energy difference required to excite the electron from \(n=1\) to \(n=3\). This energy is \(E_3 - E_1\), where \(E_1 = -13.6\) eV and \(E_3 = -13.6 \times \frac{1}{3^2}\) eV. Thus, the required energy is \(E = -13.6 \times (\frac{1}{3^2} - 1)\) eV.

Step 4: Energy Calculation

Compute \(-13.6 \times (\frac{1}{9} - 1)\) eV = 12.1 eV. This is the smallest energy needed for the \(\text{H}_{\alpha}\) emission.

Step 5: Electron Transitions from \(n=3\)

Calculate the number of possible transitions when an electron starts at \(n=3\) and ends at \(n=1\). The possible transitions are \(n=3 \to n=2\), \(n=3 \to n=1\), and \(n=2 \to n=1\).

Step 6: Wavelength Calculations for Transitions

Calculate wavelength for each transition using the Rydberg formula: \(\frac{1}{\lambda} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})\), with \(R_H \approx 1.097 \times 10^7 \text{ m}^{-1}\). Compute for \(n_3 \to n_2\), \(n_3 \to n_1\), and \(n_2 \to n_1\).

Step 7: Wavelength for Each Transition

For \(n_3 \to n_2\), calculate \(\lambda \approx 656.3\) nm. For \(n_3 \to n_1\), calculate \(\lambda \approx 102.6\) nm. For \(n_2 \to n_1\), calculate \(\lambda \approx 121.5\) nm.

Key Concepts

Balmer series

The Balmer series is a set of spectral lines corresponding to the transitions of an electron in a hydrogen atom, where the electron falls to the second energy level (\(n=2\)). These spectral lines appear in the visible spectrum, making them crucial for studying hydrogen's emission characteristics. When electrons drop from higher levels such as \(n=3, 4, 5,\) etc., to \(n=2\), they emit specific wavelengths linked to the Balmer series. It is named after Johann Balmer, who first discovered the formula that accurately predicts these wavelengths.
The most prominent line in the Balmer series is the H-alpha line, which corresponds to the transition from \(n=3\) to \(n=2\). This line is particularly important for astronomical observations because it is bright and visible, allowing astronomers to analyze stars and other celestial bodies. Understanding the Balmer series is key to grasping how hydrogen emits and absorbs energy.

Electron energy levels

Electron energy levels in an atom indicate how much energy an electron possesses in different states. In a hydrogen atom, these levels are quantized and can be described by the quantum number \(n\). The energy of an electron in a hydrogen atom is given by the formula \(E_n = -13.6 \frac{1}{n^2}\) eV. This formula tells us that the energy is negative, signifying that the electron is bound to the nucleus.
As \(n\) increases, the energy level approaches zero, meaning the electron is less tightly bound and can be easily excited to higher states. For example, the ground state (\(n=1\)) has an energy of \(-13.6\) eV, whereas higher levels like \(n=3\) have less negative energy (approximately \(-1.51\) eV), indicating greater potential energy. Transitions between these levels require absorbing or emitting precise amounts of energy, which manifests as spectral lines.

Spectral lines

Spectral lines are the result of electrons moving between different energy levels in an atom, emitting or absorbing photons in the process. Each transition corresponds to a specific amount of energy change and results in a unique spectral line with a characteristic wavelength.
When a photon is emitted as an electron falls to a lower energy level, a spectral line appears in the emission spectrum. Conversely, when an electron absorbs energy and moves to a higher level, the spectral line appears in the absorption spectrum. These lines are unique to each element, akin to fingerprints, and allow scientists to identify the elemental composition of distant stars and galaxies.
In the case of hydrogen, the spectral lines in the Balmer series appear in the visible light range, including the well-known H-alpha, H-beta, and H-gamma lines. Each line corresponds to a different electron transition, such as from \(n=3\) to \(n=2\) for the bright red H-alpha line.

Wavelength calculations

Wavelength calculations are essential to determine the electromagnetic radiation emitted or absorbed during electron transitions. The Rydberg formula is crucial for this purpose: \(\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\), where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant (approximately \(1.097 \times 10^7 \text{ m}^{-1}\)), \(n_1\) and \(n_2\) are the initial and final energy levels respectively.
By applying this formula, one can calculate the wavelength for any transition. For instance, a transition from \(n=3\) to \(n=2\) yields a wavelength of approximately 656.3 nm (H-alpha line), which falls in the red part of the visible spectrum. Similarly, the Rydberg formula helps in finding the wavelengths for transitions such as \(n=3\) to \(n=1\), resulting in ultraviolet wavelengths like 102.6 nm.
This calculation is vital for astronomers and physicists as it allows them to interpret spectral data and understand the processes occurring within stars and other cosmic bodies.