Question

(a) Consider the following three statements: (i) A hydrogen atom in the $n=3$ state can emit light at only two specific wavelengths, (ii) a hydrogen atom in the $n=2$ state is at a lower energy than the $n=1$ state, and (iii) the energy of an emitted photon equals the energy difference of the two states involved in the emission. Which of these statements is or are true? (b) Does a hydrogen atom "expand" or "contract" as it moves from its ground state to an excited state?

Short answer

(a) Statement (i) is true as a hydrogen atom in the $n=3$ state can emit light at two specific wavelengths as it transitions to the $n=1$ and $n=2$ states. Statement (ii) is false, since a hydrogen atom in the $n=2$ state has higher energy than the $n=1$ state. Statement (iii) is true, as the energy of an emitted photon is equal to the energy difference of the initial and final states. (b) A hydrogen atom "expands" as it moves from its ground state to an excited state, since its electron moves to a higher energy level further from the nucleus.

Step-by-step solution

(a) Analyzing Statement (i)

According to the Bohr model, the energy of a hydrogen atom in any given state n is given by: $$E_n=-\frac{13.6\text{eV}}{n^2}$$ When a hydrogen atom moves from a higher energy state to a lower energy state, it emits light at a specific wavelength, which can be calculated using the Rydberg formula: $$\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$$ If the hydrogen atom is in the $n=3$ state, it can move to any of the lower states (i.e., $n=1$ and $n=2$) while emitting light, meaning there are two possible values for $\lambda$. Thus, this statement is true.

(a) Analyzing Statement (ii)

According to the energy equation, the energy levels of a hydrogen atom decrease with an increase in $n$. As a result, the $n=2$ state would have a higher energy level than the $n=1$ state: $$E_1=-\frac{13.6\text{eV}}{1^2}=-13.6\text{eV}$$ $$E_2=-\frac{13.6\text{eV}}{2^2}=-3.4\text{eV}$$ Since the energy of the $n=1$ state is lower than the energy of the $n=2$ state, this statement is false.

(a) Analyzing Statement (iii)

The energy of an emitted photon should indeed equal the difference between the initial and final energy states: $$E_{\text{photon}}=E_{\text{initial}}-E_{\text{final}}$$ The photon's energy is then converted to a wavelength using the relationship: $E_{\text{photon}}=\frac{h\times c}{\lambda }$, where $h$ is Planck's constant and $c$ is the speed of light. Therefore, this statement is true.

(b) Hydrogen Atom Behavior

When a hydrogen atom moves from its ground state (i.e., the lowest energy state, $n=1$) to an excited state (i.e., higher energy state, such as $n=2$, $n=3$, etc.), its electron goes to a higher energy level, which is farther away from the nucleus. As a result, the overall size (or radius) of the atom increases. In simple terms, the hydrogen atom "expands" as it moves from its ground state to an excited state.

Key concepts

Hydrogen Atom

The hydrogen atom is the simplest and most abundant atom in the universe. It consists of just one proton in its nucleus and one electron orbiting around it. Despite its simplicity, the hydrogen atom plays a pivotal role in understanding atomic structure and quantum mechanics.

In 1913, Niels Bohr proposed a model to explain the structure of the hydrogen atom. According to the Bohr model, electrons move in circular orbits around the nucleus at fixed distances. These orbits are also known as energy levels or shells. Each of these energy levels is associated with a specific energy.

For the hydrogen atom, the energy of an electron in an energy level labeled by the quantum number "n" is given by the equation:

• $$E_n=-\frac{13.6\text{eV}}{n^2}$$

This means the electron's energy is quantized and can only have specific values, the lowest being when the electron is closest to the nucleus (called the ground state, or $n=1$).

Energy Levels

Energy levels are the different positions in which electrons can be found, relative to the nucleus in an atom. In the Bohr model, these levels are quantized and are represented by integer values like $n=1,2,3,...$, with $n=1$ being the closest to the nucleus and having the lowest energy.

The energy levels increase as the distance from the nucleus increases, meaning that electrons in higher energy levels (larger $n$ values) are less tightly bound to the nucleus. This is why the energy of these levels is expressed negatively; the closer the electron is to the nucleus, the less energy it has (more stable).

For example:

• The energy of the $n=1$ state (ground state) is $-13.6eV$.
• The energy of the $n=2$ state is $-3.4eV$.

As $n$ increases, the energy becomes less negative, illustrating that electrons require absorbing energy to jump from a lower to a higher energy level, thus moving to what is called an 'excited state.'

Photon Emission

Photon emission is a process that occurs when an electron in an atom transitions from a higher energy level to a lower one. During this transition, energy is released in the form of a photon, which is a particle of light. This is one of the fundamental processes that explains how light is emitted from atoms.

In the Bohr model, when an electron jumps down from an excited state (higher energy level) to a lower energy state, a photon with energy equal to the difference between these energy levels is emitted. The relationship can be described by the formula:

• $$E_{\text{photon}}=E_{\text{initial}}-E_{\text{final}}$$

Where $E_{\text{photon}}$ is the energy of the photon emitted, $E_{\text{initial}}$ and $E_{\text{final}}$ are the energies of the initial and final states, respectively.

The emitted photon's energy determines its wavelength (or color) using the formula:

• $$E_{\text{photon}}=\frac{h\times c}{\lambda }$$

Here, $h$ is Planck's constant and $c$ is the speed of light. Thus, the photons emitted during electronic transitions become the characteristic spectral lines of the hydrogen atom, each unique to a specific transition between energy levels.