Question

An excited hydrogen atom, whose electron is in an \(n=4\) state, is motionless. When the electron falls into the ground state, does it set the atom in motion? If so, with what speed?

Short answer

A: Yes, the hydrogen atom starts moving, and its speed is approximately \(3.63\times 10^{-2}\,\text{m/s}\).

Step-by-step solution

Calculate the energy released during the transition

The energy released when an electron falls from an initial state \(n_i\) to a final state \(n_f\) can be calculated using the following formula: \(E = -13.6\,\text{eV}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\) In our case, the initial state \(n_i = 4\) and the final state (ground state) \(n_f = 1\). Plugging these values into the formula gives: \(E = -13.6\,\text{eV}\left(\frac{1}{1^2} - \frac{1}{4^2}\right) = -13.6\,\text{eV}\left(1 - \frac{1}{16}\right)\) \(E = -13.6\,\text{eV}\left(\frac{15}{16}\right) = -12.75\,\text{eV}\) The energy released during the transition is \(12.75\,\text{eV}\).

Convert the energy to Joules

To calculate the speed, we need to convert the energy from electron volts (eV) to Joules (J). The conversion factor is \(1\,\text{eV} = 1.6\times 10^{-19}\,\text{J}\). Therefore, \(E = -12.75\,\text{eV}\times 1.6\times 10^{-19}\,\text{J/eV} = -2.04\times 10^{-18}\,\text{J}\).

Calculate the velocity of the electron

When the electron drops from the n=4 state to the ground state, it gains kinetic energy equal to the energy released during the transition. The conservation of momentum dictates that the hydrogen atom (proton) gains an equal amount of momentum but in the opposite direction. We can use the following equation to calculate the velocity of the electron: \(v_e = \sqrt{\frac{2E}{m_e}}\) where \(m_e\) is the mass of the electron (\(9.11\times10^{-31}\,\text{kg}\)). \(v_e = \sqrt{\frac{2\times -2.04\times 10^{-18}\,\text{J}}{9.11\times10^{-31}\,\text{kg}}} = 6.674\times 10^5\,\text{m/s}\)

Calculate the hydrogen atom's velocity

To calculate the velocity of the hydrogen atom, we can use the conservation of momentum. The mass of the proton (\(m_p\)) is \(1.67\times 10^{-27}\,\text{kg}\). The total initial momentum is zero. \(m_p v_p + m_e v_e = 0\) Solving for the proton's velocity (\(v_p\)): \(v_p = -\frac{m_e v_e}{m_p} = -\frac{(9.11\times10^{-31}\,\text{kg})(6.674\times 10^5\,\text{m/s})}{1.67\times 10^{-27}\,\text{kg}}\) \(v_p = 3.628\times 10^{-2}\,\text{m/s}\) The hydrogen atom will be set in motion, and its speed will be approximately \(3.63\times 10^{-2}\,\text{m/s}\).

Key concepts

Energy Levels

Energy levels in a hydrogen atom are like rungs on a ladder that the electron can "climb." Each level is characterized by a principal quantum number, denoted as \( n \). In the original exercise, the electron transitions from an energy level where \( n=4 \), meaning it starts from the fourth rung from the bottom, to \( n=1 \), the lowest level, also known as the ground state.

Energy levels are quantized; that is, an electron can only exist in specific energy states. The energy of each state can be represented by the formula:

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

For each transition an electron makes between these levels, a specific amount of energy is released in the form of electromagnetic radiation, commonly seen as light.

The energy difference between levels \( n_i \) and \( n_f \) can be calculated by the difference in their energies, which gives us:

• \( E = -13.6 \, \text{eV} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \)

When applying this to our hydrogen transition from \( n=4 \) to \( n=1 \), the calculation gives the energy released as \( 12.75 \, \text{eV} \). Understanding these energy differences helps explain why specific colors of light are emitted during atomic transitions.

Conservation of Momentum

The law of conservation of momentum states that the total momentum of a closed system is constant if it is not affected by external forces. This principle is critical in understanding why the hydrogen atom in the exercise is set in motion.

Consider that before the transition, the system is motionless (total momentum is zero). Upon the electron's move to a lower energy state, it emits a photon carrying away momentum. To keep the total momentum of the system unchanged, the hydrogen atom (or proton) responds by moving in the opposite direction.

The resulting change in movement ensures balance, adhering to the principle:

• \( m_p v_p + m_e v_e = 0 \)

Where \( m_p \) and \( m_e \) are the masses of the proton and electron, respectively, and \( v_p \) and \( v_e \) their velocities.

The negative sign implies that their momenta are in opposite directions. The conservation principle allows calculation of the hydrogen atom's velocity post-transition, showing it, too, gains speed as the system compensates for the energy released.

Velocity Calculations

Once the transition energy is calculated and understood, you can determine the velocities for both the electron and the hydrogen atom using some basic physics principles.

Firstly, to find the velocity of the electron, we use kinetic energy and its relation to velocity. The kinetic energy which the electron gains is equal to the transition energy, so:

• \( v_e = \sqrt{\frac{2E}{m_e}} \)

Given that the energy \( E \) is approximately \(-2.04 \times 10^{-18} \, \text{J} \) and the electron's mass \( m_e \) is \( 9.11\times10^{-31} \, \text{kg} \), the electron's velocity works out to be about \( 6.674 \, \times 10^5 \, \text{m/s} \).

Next, using conservation of momentum, the hydrogen atom's velocity can be calculated. Seeing the proton's mass \( m_p \) is much larger than the electron's, its velocity is much smaller, found using:

• \( v_p = -\frac{m_e v_e}{m_p} \)

Plug in the provided values to get \( v_p = 3.628\times 10^{-2} \, \text{m/s} \). This final velocity translation reflects the principles of momentum conservation, highlighting how a system adjusts to maintain balance.