Problem 15 - College Physics

Question

(a) How much energy is needed to ionize a hydrogen atom that is in the \(n=4\) state? (b) What would be the wavelength of a photon emitted by a hydrogen atom in a transition from the \(n=4\) state to the \(n=2\) state?

Step-by-Step Solution

Answer

(a) 0.85 eV, (b) 486 nm

Step 1: Determine the Energy of Ionization

For a hydrogen atom in the nth state, the energy needed for ionization can be determined using the formula for energy levels \(E_n = -13.6 \text{ eV} \times \frac{1}{n^2}\). To ionize, the atom needs to reach zero energy. Therefore, the energy required to ionize a hydrogen atom from the state \(n=4\) is \(-E_4\). Calculate \(E_4 = -13.6 \times \frac{1}{4^2} = -13.6 \times \frac{1}{16} = -0.85 \text{ eV}\). Hence, the energy needed is \(0.85 \text{ eV}\).

Step 2: Calculate the Energy of Photon Emission

The energy of the photon emitted during the transition from \(n=4\) to \(n=2\) can be found using the difference in energy levels \(E = E_2 - E_4\). Calculate \(E_2 = -13.6 \times \frac{1}{2^2} = -3.4 \text{ eV}\) and already know \(E_4 = -0.85 \text{ eV}\). Thus, \(E = -3.4 - (-0.85) = -2.55 \text{ eV}\).

Step 3: Determine the Wavelength of the Emitted Photon

Use the energy-wavelength relation \(E = \frac{hc}{\lambda}\) to find the wavelength, where \(h\) is Planck's constant \(4.1357 \times 10^{-15} \text{ eV}\cdot\text{s}\) and \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\). Rearrange the formula to find \(\lambda = \frac{hc}{E}\). Substitute \(E = 2.55 \text{ eV}\). Convert \(E\) to Joules: \(2.55 \text{ eV} = 2.55 \times 1.602 \times 10^{-19} \text{ J}\). Calculate \(\lambda = \frac{(4.1357 \times 10^{-15})(3 \times 10^8)}{2.55 \times 1.602 \times 10^{-19}} \approx 486 \text{ nm}\).

Key Concepts

Energy Levels

Energy levels are specific energies that electrons can have in an atom. These levels are quantized, meaning electrons can only occupy certain allowed energy states. For a hydrogen atom, these levels are calculated using the formula: \[ E_n = -13.6 ext{ eV} \times \frac{1}{n^2} \] where \( n \) is the principal quantum number. The negative sign indicates that the electron is bound to the nucleus; the closer \( n \) is to 1, the lower and more negative the energy. - **Ground state**: The lowest energy level \( (n=1) \). - **Excited states**: Higher energy levels \( (n>1) \). Each level corresponds to a different orbit of the electron around the nucleus. When an electron transitions between these levels, energy is absorbed or emitted.

Photon Emission

Photon emission occurs when an electron moves from a higher energy level to a lower one. In this process, energy is released in the form of a photon, a particle of light. The amount of energy released corresponds to the difference in energy between the two levels. For example, when an electron transitions from \( n=4 \) to \( n=2 \) in a hydrogen atom: 1. Calculate the energy at both levels: \[ E_4 = -0.85 ext{ eV}, \] \[ E_2 = -3.4 ext{ eV} \]2. Determine the energy difference: \[ E = E_2 - E_4 = -2.55 ext{ eV} \]This energy difference represents the energy of the emitted photon. The photon carries this energy as it leaves the atom.

Wavelength of Photon

The wavelength of a photon is inversely related to its energy. This means a smaller wavelength corresponds to a higher energy photon. The energy-wavelength relationship is given by: \[ E = \frac{hc}{\lambda} \] where: - \( E \) is energy in Joules, - \( h \) is Planck's constant \( (4.1357 \times 10^{-15} \text{ eV}\cdot\text{s}) \), - \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \), - \( \lambda \) is the wavelength. By rearranging the formula, we can solve for wavelength: \[ \lambda = \frac{hc}{E} \] In our example, substituting the emitted photon energy, 2.55 eV, gives us a wavelength of approximately 486 nm. This wavelength typically falls in the visible spectrum.

N-State Transition

In quantum physics, \(n\)-state transitions refer to an electron moving between two energy levels of a hydrogen atom. Transitions are often labeled as \( \Delta n \), showing a change from one principal quantum number to another. For hydrogen, the transition from a higher state to a lower state releases a photon. The energy of this photon is given by the change in energy levels: \[ E = E_{initial} - E_{final} \] Significant \(n\)-state transitions include: - **Balmer Series**: Transitions into \( n=2 \) produce photons in visible light. Example: Free fall from \( n=4 \) to \( n=2 \). - **Lyman Series**: Transitions into \( n=1 \) emit ultraviolet light. - **Paschen Series**: Transitions into \( n=3 \) are found in infrared.Understanding these transitions helps explain phenomena like spectral lines and the atom's energy spectrum.