Problem 39 - Chemistry: The Central Science

Question

(a) Using Equation 6.5, calculate the energy of an electron in the hydrogen atom when \(n=2\) and when \(n=6\). Calculate the wavelength of the radiation released when an electron moves from \(n=6\) to \(n=2\). (b) Is this line in the visible region of the electromagnetic spectrum? If so, what color is it?

Step-by-Step Solution

Answer

The energy levels of the electron in the hydrogen atom at \(n = 2\) and \(n = 6\) are approximately \(-3.4 eV\) and \(-0.3778 eV\), respectively. The energy difference between these levels is about \(-3.0222 eV\), and the wavelength of the emitted radiation during the transition is \(410 nm\), which corresponds to violet light within the visible spectrum.

Step 1: Identify the given equation and constants

Equation 6.5 refers to the equation for the energy of an electron in a hydrogen atom, given by: \[E_n = -\frac{13.6 eV}{n^2}\] Here, \(E_n\) is the energy of the electron at the nth energy level and \(n\) is an integer representing the energy level. We are given \(n = 2\) and \(n = 6\) as the energy levels for this problem.

Step 2: Calculate the energy of the electron at n = 2 and n = 6

Using the given equation, calculate the energy of an electron for both states: For \(n = 2\): \[E_2 = -\frac{13.6 eV}{2^2} = -\frac{13.6 eV}{4} = -3.4 eV\] For \(n = 6\): \[E_6 = -\frac{13.6 eV}{6^2} = -\frac{13.6 eV}{36} = -0.3778 eV\]

Step 3: Calculate the energy difference between the two states

To calculate the energy released when the electron transitions from n = 6 to n = 2, subtract the final energy level (\(E_2\)) from the initial energy level (\(E_6\)): \[ΔE = E_2 - E_6 = -3.4 eV -(-0.3778 eV) = -3.0222eV\]

Step 4: Calculate the wavelength of radiation released during the transition

Use the following relationship to relate energy and wavelength: \[E = \frac{hc}{λ}\] Where \(E\) is the energy difference, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} Js\)), \(c\) is the speed of light (\(3 \times 10^8 m/s\)), and \(λ\) is the wavelength. Solve for \(λ\): \[λ = \frac{hc}{E}\] First, convert the energy difference (ΔE) from electron volts to joules by multiplying it by the electron charge: \[-3.0222 eV \times \frac{1.602 \times 10^{-19} J}{1 eV} = -4.84 \times 10^{-19} J\] Now, calculate the wavelength: \[λ = \frac{(6.626 \times 10^{-34} Js)(3 \times 10^8 m/s)}{-4.84 \times 10^{-19} J} = 4.1 \times 10^{-7} m = 410 nm\]

Step 5: Determine if the wavelength is within the visible spectrum and identify its color

The visible light spectrum ranges from approximately 400nm (violet) to 700nm (red). Since the calculated wavelength of 410nm falls within this range, the radiation released is in the visible spectrum. The wavelength of 410 nm corresponds to violet light. Thus, during the transition from n = 6 to n = 2, the electron emits a violet light photon.

Key Concepts

Energy Levels

In the Bohr model of the hydrogen atom, energy levels are specific orbits around the nucleus where electrons reside. These are quantized, meaning that electrons can only exist at certain energy levels, which are characterized by the principal quantum number, \( n \). Each energy level has a distinct energy, calculated using the formula:\[E_n = -\frac{13.6 \text{ eV}}{n^2}\]Here, \( E_n \) represents the energy of the electron in the \( n \)-th energy level, where \( n \) is an integer. The negative sign indicates that electrons are bound and require energy to be moved to higher levels or removed from the atom entirely. Understanding these discrete energy levels is crucial as they determine how atoms absorb or emit radiation when electrons transition between them.

Electron Transitions

Electron transitions occur when an electron moves from one energy level to another. This can happen when an atom absorbs or emits energy, often in the form of light. When an electron falls from a higher energy level to a lower one, it releases energy, typically as a photon, which is a particle of light. For example, when an electron drops from \( n=6 \) to \( n=2 \), it moves to a lower energy state, releasing a photon carrying the energy difference between these two levels. The formula to find this energy difference is:\[ \Delta E = E_{\text{lower}} - E_{\text{higher}} \]This energy corresponds to the energy of the photon emitted. Such transitions are key to understanding atomic spectra, leading to emission lines characteristic of each element.

Visible Spectrum

The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. It ranges approximately from 400 nm (violet) to 700 nm (red). When an electron transition releases energy that falls within this range, it results in visible light, which is part of what we can observe as the colors of the hydrogen emission spectrum.In electron transitions like from \( n=6 \) to \( n=2 \), if the released photon's wavelength falls within this visible range, it will appear as a specific color. Given that the calculated wavelength for our example is 410 nm, it falls within the visible spectrum and appears as violet light. This is an example of how certain electron transitions result in distinct colors, forming a line spectrum specific to hydrogen.

Wavelength Calculation

Calculating the wavelength of radiation emitted during an electron transition is essential for connecting energy changes with observable light. The relationship between energy \( E \) and wavelength \( \lambda \) is given by:\[ E = \frac{hc}{\lambda} \]where:

• \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ Js})\).
• \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\).

By rearranging the formula, the wavelength \( \lambda \) can be determined:\[ \lambda = \frac{hc}{E} \]First, convert the energy from electron volts to joules since the constant values are in SI units. Then, use the energy difference obtained from the electron transition to compute the wavelength. For the transition from \( n=6 \) to \( n=2 \), the calculated wavelength is 410 nm, fitting the visible light criteria. This capability to measure and predict wavelengths helps understand atomic behavior and light's interaction with matter.