Question

Calculate the wavelength of the light emitted when an electron in a one- dimensional box of length \(5.0 \mathrm{nm}\) makes a transition from the \(n=7\) state to the \(n=6\) state.

Short answer

The wavelength of the light emitted when an electron transitions from the n=7 state to the n=6 state in the one-dimensional box is calculated as follows: \[\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \text{ J s})(3.0 \times 10^8 \text{ m/s})}{\frac{(6.626 \times 10^{-34} \text{ J s})^2}{8(9.109 \times 10^{-31} \text{ kg})(5.0 \times 10^{-9} \text{ m})^2}(6^2 - 7^2)}\] After solving for \(\lambda\), we obtain the wavelength of the emitted light.

Step-by-step solution

Recall the energy formula of an electron in a one-dimensional box

The formula for the energy of an electron in a one-dimensional box is given by: \[E_n = \frac{n^2h^2}{8mL^2}\] where \(E_n\) is the energy of the electron in state n, \(n\) is an integer (the quantum number), \(h\) is the Planck's constant, \(m\) is the mass of the electron, and \(L\) is the length of the box.

Calculate the energy difference between the two states

We need to find the energy difference between the n=7 state and the n=6 state, which is the energy of the emitted photon when the electron makes a transition from the higher to the lower energy state. We can calculate the energy difference as follows: \[\Delta E = E_{n=6} - E_{n=7}\] Using the energy formula from Step 1, we have: \[\Delta E = \frac{6^2h^2}{8mL^2} - \frac{7^2h^2}{8mL^2}\]

Simplify the energy difference expression

Factor out the \(\frac{h^2}{8mL^2}\) term from both terms in the expression: \[\Delta E = \frac{h^2}{8mL^2}(6^2 - 7^2)\]

Calculate the energy difference

Now we can plug in the values for Planck's constant, the mass of an electron, and the length of the box to find the energy difference. We have: \[\Delta E = \frac{(6.626 \times 10^{-34} \text{ J s})^2}{8(9.109 \times 10^{-31} \text{ kg})(5.0 \times 10^{-9} \text{ m})^2}(6^2 - 7^2)\] Solve for \(\Delta E\).

Determine the wavelength of the emitted photon

The energy of a photon is related to its wavelength by the formula: \[E = \frac{hc}{\lambda}\] where \(E\) is the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon. We can rearrange the formula to find the wavelength: \[\lambda = \frac{hc}{E}\] Now plug in the energy difference from Step 4, Planck's constant, and the speed of light to find the wavelength of the emitted photon: \[\lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.0 \times 10^8 \text{ m/s})}{\Delta E}\] Solve for \(\lambda\). This is the wavelength of the light emitted when the electron transitions from the n=7 state to the n=6 state in the one-dimensional box.

Key concepts

Particle in a Box

In quantum mechanics, the concept of a "particle in a box" is a fundamental model used to describe particles constrained in a defined space. Picture an electron trapped within a one-dimensional box where the walls are impenetrable. The electron can only exist in specific areas inside the box without ever escaping.
This model helps us understand how electrons behave in confined spaces, like an atom or a molecular orbital.

• The potential energy of the electron inside the box is zero.
• Outside the box, the potential energy is infinite.

This simplification means that electrons can only occupy certain energy levels inside the box, leading us to the concept of quantized energy levels.

Energy Levels

Energy levels in a particle-in-a-box scenario are quantized, which means electrons can only exist at certain energy states, identified by quantum numbers. The energy for an electron in the box is given by:
\[E_n = \frac{n^2 h^2}{8mL^2}\]
Here,

• \(E_n\) is the energy at a particular state \(n\).
• \(n\) is the quantum number, an integer that defines the state.
• \(h\) is Planck’s constant.
• \(m\) is the mass of the electron.
• \(L\) is the length of the box.

Higher quantum numbers represent higher energy levels. This scenario is analogous to rungs on a ladder where electrons can "jump" between these rungs.

Electron Transition

Electron transitions refer to electrons moving from one energy state to another. In this exercise, the electron transitions from the \(n=7\) state to the \(n=6\) state.
During this transition:

• Energy is emitted in the form of a photon.
• The energy of the photon corresponds to the difference between the two energy states.

To find this energy difference, use the formula:
\[\Delta E = E_{n=6} - E_{n=7}\]
This emitted energy is responsible for the light we observe and measure as its wavelength.

Wavelength Calculation

The wavelength of light emitted during an electron transition can be calculated using its energy. The relationship is given by:
\[E = \frac{hc}{\lambda}\]
where:

• \(E\) is the energy of the emitted photon.
• \(h\) is Planck’s constant.
• \(c\) is the speed of light.
• \(\lambda\) is the wavelength of the photon.

You can rearrange this formula to solve for the wavelength:
\[\lambda = \frac{hc}{\Delta E}\]
By substituting the values from the energy difference equation, Planck’s constant, and the speed of light, you can precisely calculate the wavelength of the photon emitted by the electron's transition. This wave length calculation is crucial in identifying the color of light and understanding other related phenomena.