Question

Classify each of the following statements as either true or false: (a) A hydrogen atom in the \(n=3\) state can emit light at only two specific wavelengths, \((\mathbf{b})\) a hydrogen atom in the \(n=2\) state is at a lower energy than one in the \(n=1\) state, and (c) the energy of an emitted photon equals the energy difference of the two states involved in the emission.

Short answer

Based on the analysis of given statements: (a) This statement is \(\mathbf{True}\), because an electron in the n=3 state can emit light at only two specific wavelengths. (b) This statement is \(\mathbf{False}\), as a hydrogen atom in the n=1 state is at lower energy than one in the n=2 state. (c) This statement is \(\mathbf{True}\), as the energy of an emitted photon equals the energy difference of the two states involved in the emission.

Step-by-step solution

Statement (a) Classification

We are asked to determine if a hydrogen atom in the n=3 can emit light at only two specific wavelengths. To analyze this, we have to recall that when an electron transitions between states, the energy differences correspond to specific photon energies, which in turn correspond to specific wavelengths. When an electron in the n=3 state transitions to a lower-energy state, the possible lower-energy states are n=2 and n=1. So, we have: - Transition from n=3 to n=2 - Transition from n=3 to n=1 Hence, an electron in the n=3 state can emit light at only two specific wavelengths. This statement is therefore \(\mathbf{True}\).

Statement (b) Classification

We are asked to determine if a hydrogen atom in the n=2 state is at a lower energy than one in the n=1 state. To analyze this, we have to recall the formula for the energy levels of hydrogen state, which is given by: \[ E_n = - \frac{13.6 eV}{n^2} \] Where n is the state, and E is the energy level. For n=1, we have: \(E_1 = - \frac{13.6 eV}{1^2} = -13.6 eV \) For n=2, we have: \(E_2 = - \frac{13.6 eV}{2^2} = -\frac{13.6 eV}{4} = -3.4 eV \) Since -13.6 eV is less than -3.4 eV, a hydrogen atom in the n=1 state is at lower energy than one in the n=2 state. This statement is therefore \(\mathbf{False}\).

Statement (c) Classification

We are asked to determine if the energy of an emitted photon equals the energy difference of the two states involved in the emission. To analyze this, we have to recall that when an electron transitions from a higher energy state to a lower energy state, it emits a photon with the exact energy difference between the two states. The energy difference is given by: \( \Delta E = E_\text{final} - E_\text{initial} \) Since the energy of the emitted photon equals the energy difference between the two states, this statement is therefore \(\mathbf{True}\).

Key concepts

Electron Transitions

Electron transitions are fascinating events that occur when electrons move between different energy levels within an atom. In the hydrogen atom, electrons can jump between discrete energy levels, which are essentially the allowed paths for the electron around the nucleus. When an electron transitions from a higher energy level to a lower one, it releases energy.

- An electron moving from a higher energy state (say, n=3) to a lower energy state (such as n=2 or n=1) will emit energy in the form of light. - These energy transitions correspond to specific wavelengths of light that are emitted as photons.

The transitions aren't random; they follow certain rules: only certain transitions are allowed, and each one involves a specific amount of energy. This energy difference dictates the frequency and wavelength of the emitted photon, establishing a direct relationship between electron transitions and photon emission.

Energy Levels in Hydrogen

The energy levels of the hydrogen atom are quantized, meaning that electrons can only occupy certain energy levels around the nucleus. This is described by the formula:\[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \]
Here, \(n\) is the principal quantum number, which takes positive integer values (1, 2, 3,...), directly related to the energy level. The negative sign indicates that energy is lower than that of a free electron (which would have zero energy).

- **For n=1:** The energy level is \(-13.6 \, \text{eV}\), which is the lowest and most stable energy level.- **For n=2:** The energy level is \(-3.4 \, \text{eV}\), indicating a higher energy but less stability compared to \(n=1\).

These calculated energy values show that energy becomes less negative (and thus higher) as the electron moves to higher levels. Electrons tend to occupy the lowest available energy level, explaining why a hydrogen atom in the \(n=1\) state is more stable than in the \(n=2\) state.

Photon Emission

When an electron in a hydrogen atom transitions to a lower energy level, it emits energy in the form of a photon. This is a crucial concept for understanding how light is produced at the atomic level.

- **Energy Difference**: The energy of the photon emitted exactly equals the energy difference between the initial and final states.- The formula for the energy difference is \[ \Delta E = E_{\text{final}} - E_{\text{initial}} \]

This emitted photon carries a specific amount of energy, found using the formula:\[ E = h u \]where \(E\) is the energy of the photon, \(h\) is Planck's constant, and \(u\) is the frequency of the light. This relationship shows how changes at the microscopic level, like electron transitions, result in observable phenomena, such as the emission of light at particular wavelengths.