Question

For each of the following electronic transitions in the hydrogen atom, calculate the energy, frequency, and wavelength of the associated radiation, and determine whether the radiation is emitted or absorbed during the transition: (a) from \(n=4\) to \(n=1\), (b) from \(n=5\) to \(n=2,(\mathrm{c})\) from \(n=3\) to \(n=6\). Does any of these transitions emit or absorb visible light?

Short answer

The electronic transitions in a hydrogen atom are as follows: a) from n=4 to n=1 emits radiation with energy -12.75 eV, frequency 3.083 x 10¹⁵ Hz, and wavelength 97.3 nm, which is not visible; b) from n=5 to n=2 emits radiation with energy -2.73 eV, frequency 6.593 x 10¹⁴ Hz, and a visible wavelength of 455 nm; c) from n=3 to n=6 absorbs radiation with energy 1.13 eV, frequency 2.733 x 10¹⁴ Hz, and wavelength 1098 nm, which is not visible.

Step-by-step solution

Energy Calculation

Calculate the energy difference between the initial and final state using ΔE equation: \( \Delta E = (-13.6 \ \text{eV}) \left(\frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2}\right) \) \( \Delta E = (-13.6 \ \text{eV}) \left(\frac{1}{1^2} - \frac{1}{4^2}\right) \) \( \Delta E = -13.6 (1 - 0.0625) \ \text{eV} \) \( \Delta E = -12.75 \ \text{eV} \) Since ΔE is negative, the energy is emitted during the transition.

Frequency Calculation

Calculate the frequency of the emitted radiation using the frequency equation: ν = ΔE/h ν = (-12.75 eV)/(4.13567 x 10⁻¹⁵ eV/s) ν = 3.083 x 10¹⁵ Hz

Wavelength Calculation

Calculate the wavelength of the emitted radiation using the wavelength equation: \( \lambda = \frac{c}{\nu} \) \( \lambda = \frac{3 \cdot 10^8 \ \text{m/s}}{3.083 \cdot10^{15} \ \text{Hz}} \) \( \lambda = 9.73 \cdot 10^{-8} \ \text{m} \) \( \lambda = 97.3 \ \text{nm} \) Since the wavelength is outside the range of visible light (400 nm to 700 nm), this transition doesn't emit visible light. #b) Transition from n=5 to n=2#

Energy Calculation

Calculate the energy difference between the initial and final state using ΔE equation: \( \Delta E = (-13.6 \ \text{eV}) \left(\frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2}\right) \) \( \Delta E = (-13.6 \ \text{eV}) \left(\frac{1}{2^2} - \frac{1}{5^2}\right) \) \( \Delta E = -13.6 (0.25 - 0.04) \ \text{eV} \) \( \Delta E = -2.73 \ \text{eV} \) Since ΔE is negative, the energy is emitted during the transition.

Frequency Calculation

Calculate the frequency of the emitted radiation using the frequency equation: ν = ΔE/h ν = (-2.73 eV)/(4.13567 x 10⁻¹⁵ eV/s) ν = 6.593 x 10¹⁴ Hz

Wavelength Calculation

Calculate the wavelength of the emitted radiation using the wavelength equation: \( \lambda = \frac{c}{\nu} \) \( \lambda = \frac{3 \cdot 10^8 \ \text{m/s}}{6.593 \cdot10^{14} \ \text{Hz}} \) \( \lambda = 4.55 \cdot 10^{-7} \ \text{m} \) \( \lambda = 455 \ \text{nm} \) Since the wavelength is within the range of visible light (400 nm to 700 nm), this transition emits visible light. #c) Transition from n=3 to n=6#

Energy Calculation

Calculate the energy difference between the initial and final state using ΔE equation: \( \Delta E = (-13.6 \ \text{eV}) \left(\frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2}\right) \) \( \Delta E = (-13.6 \ \text{eV}) \left(\frac{1}{6^2} - \frac{1}{3^2}\right) \) \( \Delta E = -13.6 (0.0278 - 0.1111) \ \text{eV} \) \( \Delta E = 1.13 \ \text{eV} \) Since ΔE is positive, the energy is absorbed during the transition.

Frequency Calculation

Calculate the frequency of the absorbed radiation using the frequency equation: ν = ΔE/h ν = (1.13 eV)/(4.13567 x 10⁻¹⁵ eV/s) ν = 2.733 x 10¹⁴ Hz

Wavelength Calculation

Calculate the wavelength of the absorbed radiation using the wavelength equation: \( \lambda = \frac{c}{\nu} \) \( \lambda = \frac{3 \cdot 10^8 \ \text{m/s}}{2.733 \cdot10^{14} \ \text{Hz}} \) \( \lambda = 1.098 \cdot 10^{-6} \ \text{m} \) \( \lambda = 1098 \ \text{nm} \) Since the wavelength is outside the range of visible light (400 nm to 700 nm), this transition doesn't absorb visible light. In conclusion, the transitions a) and b) emit radiation while transition c) absorbs radiation. Only transition b) emits visible light.

Key concepts

Energy Calculation in Electronic Transitions

Understanding the energy changes during an electronic transition in a hydrogen atom is pivotal for grasping the nature of atomic processes. The key principle here is the conservation of energy. Simply put, when an electron moves between energy levels, the difference in energy between the higher energy level (n_initial) and the lower energy level (n_final) is either emitted (in case of a drop down to a lower energy level) or absorbed (when jumping up to a higher energy level).

To calculate this energy change (\Delta E), we use the formula: \[\Delta E = (-13.6 \text{ eV}) \left(\frac{1}{{n_{{final}}^2}} - \frac{1}{{n_{{initial}}^2}}\right)\]. The negative sign in front of the 13.6 eV indicates that energy levels become more negative as the electron moves to higher energy levels, which is a convention in physics for the potential energy in the electrostatic field of the nucleus.

The calculation of \Delta E shows if the energy will be released or absorbed during a transition. If \Delta E is negative, the transition results in the emission of a photon, carrying away energy. Conversely, a positive \Delta E implies that energy must be absorbed from an incoming photon to enable the transition to the higher energy level.

Wavelength and Frequency Determination

Once the energy of an electron transition has been calculated, the next steps are to determine the frequency and wavelength of the associated radiation. These are key concepts in understanding how energy is quantified and observed at the atomic level.

The frequency (u) of the radiation represents how many waves pass a point each second and is calculated using Planck's constant (h): \[u = \frac{\Delta E}{h}\]. Directly related to frequency is the wavelength (ambda), which is the distance between identical points in the adjacent cycles of a waveform. The wavelength can be found with the speed of light (c) and frequency: \[ambda = \frac{c}{u}\].

This relationship between energy, frequency, and wavelength is fundamental in quantum mechanics and understanding the behavior of photons and electronic transitions in atoms. These calculations let us determine whether the emitted or absorbed radiation lies within the spectrum of visible light or if it falls within a different range, such as ultraviolet or infrared.

Visible Light Emission and Absorption

The visible light spectrum is a small part of the electromagnetic spectrum that the human eye can detect, typically ranging from 400 nm (violet) to 700 nm (red). When an electron undergoes a transition within this range, the resulting photon can be seen as colored light. If the transition falls outside of this range, it is invisible to us, though still detectable with instruments.

From the exercise provided, we calculated the wavelength of the radiation involved in various transitions within the hydrogen atom. We can determine whether the radiation is part of the visible spectrum by comparing the resulting wavelength to the known limits of human vision. For example, a transition found to result in a wavelength of 455 nm would emit a visible blue light. In contrast, wavelengths shorter than 400 nm or longer than 700 nm, need other types of equipment than the human eye to be detected, as they lie in the ultraviolet or infrared regions respectively.

The understanding of which transitions result in visible light emission or absorption is not only fascinating but also essential in fields such as spectroscopy, where this knowledge is used to analyze the composition of substances or the astronomical observation of distant stars to determine their properties based on the spectral lines emitted by their constituent atoms.