Question

The electron in a hydrogen atom relaxes from the \(n=4\) shell to some lower- energy shell. The light emitted during the relaxation has a wavelength of \(1772.6 \mathrm{~nm}\). By calculating the energy of this light, determine the shell to which the electron relaxed. \(\left[1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\right]\)

Short answer

The energy of the emitted light is calculated as \(E = \frac{6.626 \times 10^{-34} J \cdot s \cdot 3 \times 10^8 m/s}{1772.6 \times 10^{-9} m}\). After converting the energy to eV, we use the Rydberg formula: \(\frac{1}{\lambda_{vac}} = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\) to find the final shell \(n_f^2 = \left(\frac{1}{4^2} - \frac{1772.6 \times 10^{-9}}{1.097373 \times 10^{7}}\right)^{-1}\). Then, \(n_f = \sqrt{n_f^2}\), and round it to the nearest whole number. The electron relaxed to shell \(n_f = 2\).

Step-by-step solution

Calculate the energy of emitted light

First, we need to calculate the energy of the light emitted during the relaxation. The energy can be calculated using the formula E = h * c / λ, where h = 6.626 x 10^{-34} J s (Planck's constant), c = 3 x 10^8 m/s (speed of light), and λ = 1772.6 nm = 1772.6 x 10^{-9} m (wavelength). E = \(\frac{6.626 \times 10^{-34} J \cdot s \cdot 3 \times 10^8 m/s}{1772.6 \times 10^{-9} m}\)

Convert energy to eV

Now we need to convert the energy calculated in Step 1 from Joules to electron volts (eV). The conversion factor is given: [1 eV \( = 1.602 \times 10^{-19} J\)]. To convert the energy, divide it by the conversion factor: E_eV = \(\frac{E}{1.602 \times 10^{-19}}\)

Use the Rydberg formula to find the final shell

Now, we will use the Rydberg formula to find the final shell (n_f): \(\frac{1}{\lambda_{vac}} = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\) where λ_vac is the vacuum wavelength (given), R_H is the Rydberg constant for hydrogen \(\left(1.097373 \times 10^{7} m^{-1}\right)\), n_i is the initial shell (\(n_i = 4\)), and n_f is the final shell (unknown). Rearrange the formula to solve for n_f: \(n_f^2 = \left(\frac{1}{n_i^2} - \frac{\lambda_{vac}}{R_H}\right)^{-1}\) Plug in the known values: \(n_f^2 = \left(\frac{1}{4^2} - \frac{1772.6 \times 10^{-9}}{1.097373 \times 10^{7}}\right)^{-1}\) Calculate the final shell (n_f) by taking the square root: \(n_f = sqrt(n_f^2)\)

Determine the shell to which the electron relaxed

Now that we have calculated n_f, round it to the nearest whole number to find the shell to which the electron relaxed.

Key concepts

Electron Relaxation

Electron relaxation is a fascinating process that occurs when an electron in an atom moves from a higher energy state, or excited state, to a lower energy state. This transition releases energy in the form of light, known as a photon. In the case of a hydrogen atom, an electron may move from a higher shell such as the fourth shell to a lower shell. This movement is driven by the electron's tendency to occupy the lowest possible energy state, resulting in light emission.
In the exercise, the electron starts in the fourth energy level (n=4). As it relaxes, it emits light with a specific wavelength. This information helps us determine its final resting shell. By understanding electron relaxation, we can explore how light relates to changes in the energy levels of atoms.

Rydberg Formula

The Rydberg formula is essential for calculating the wavelengths of light emitted or absorbed as electrons transition between energy levels in hydrogen atoms. It’s a handy equation that relates the initial and final energy levels of the electron to the wavelength of the emitted light.
The formula is expressed as: \[ \frac{1}{\lambda_{vac}} = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right) \] where \(\lambda_{vac}\) is the wavelength, \(R_H\) is the Rydberg constant for hydrogen, \(n_i\) is the initial energy level, and \(n_f\) is the final energy level.
Using this formula, we can solve for the unknown final shell \(n_f\) by rearranging the formula and inserting known values. This calculation effectively uses the properties of the photon emitted during electron relaxation to infer the electron’s skydiving endpoint in the atom's energy stairs.

Energy and Wavelength Conversion

Energy and wavelength conversion is a crucial step in deriving information from observed photon emissions. Light emitted from electron relaxation comes with a specific wavelength, which can be used to calculate energy using Planck's equation: \[ E = \frac{h \cdot c}{\lambda} \] where \(E\) is energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.
Once we have the energy in joules, it's often useful to convert it to electron volts (eV) for more practical applications in atomic physics. This conversion is done using the equation: \[ E_{eV} = \frac{E}{1.602 x 10^{-19}} \] This step ensures that we can relate the photon’s energy to the electron transitions described by the Rydberg formula, seamlessly converting physical observations into meaningful quantum-level insights.

Hydrogen Emission Spectrum

The hydrogen emission spectrum is a set of wavelengths emitted by the transitions of electrons within hydrogen atoms. When electrons drop to lower energy levels, they release specific amounts of light, contributing to this spectrum. The spectrum is composed of distinct lines, each corresponding to a different electronic transition.
These lines form series, such as the Lyman or Balmer series, depending on the final energy level of the electron. The Balmer series, for example, appears in the visible light range.
Understanding the hydrogen emission spectrum helps scientists identify and analyze the atomic structure and behavior of hydrogen. Besides, it plays a pivotal role in discerning how atoms of different elements interact with light in various physical and chemical processes. This aspect of atomic physics underscores the importance of studying atomic electron transitions and their observable manifestations in the spectral world.