Question

Calculate the highest possible energy of a photon that can be observed in the emission spectrum of \(\mathrm{H}\).

Short answer

The highest possible energy of a photon that can be observed in the emission spectrum of hydrogen is \(13.6 \ eV\). This corresponds to a transition from the ground state (n = 1) to an infinitely high energy level (n = ∞).

Step-by-step solution

Energy levels of hydrogen

In a hydrogen atom, the energy levels are given by the formula: \[E_n = -\frac{13.6 \ eV}{n^2}\] where \(E_n\) is the energy of the nth level (n = 1, 2, 3, ...), and eV is the unit of energy (electronvolt).

Determine the highest possible energy difference

The highest possible energy difference would correspond to a transition from the highest energy level (n = 1, the ground state) to the lowest energy level (n = ∞, very far from the nucleus). So, we must calculate the energy difference between these two levels: \[E_{diff} = E_{\infty} - E_1\]

Calculate the energy difference between the levels

Using the formula for energy levels, we can find the energy of the first and "infinite" levels: \[E_1 = -\frac{13.6 \ eV}{1^2} = -13.6 \ eV\] \[E_{\infty} = -\frac{13.6 \ eV}{(\infty)^2} = 0 \ eV\] Now, substitute these values into the energy difference formula: \[E_{diff} = 0 \ eV - (-13.6 \ eV) = 13.6 \ eV\]

Conclusion

Therefore, the highest possible energy of a photon that can be observed in the emission spectrum of hydrogen is 13.6 eV.

Key concepts

Energy Levels of Hydrogen

Understanding the energy levels of hydrogen is fundamental to study and interpret the emission spectrum of a hydrogen atom. The energy levels of hydrogen are quantized, meaning the electron can only inhabit certain discrete energy levels. The formula to determine the energy of a particular level 'n' in a hydrogen atom is given by:

\[\begin{equation}E_n = -\frac{13.6 \ eV}{n^2}\end{equation}\]
Here, the energy level () corresponds to the principal quantum number, which can be any positive integer (n=1, 2, 3, ...). The negative sign indicates that these energy levels are bound states, meaning that energy must be added to the electron to free it from the atom. The unit of energy used in this formula is the electronvolt (eV), which is a unit of energy that's particularly convenient when dealing with atomic-scale phenomena.
The ground state, or the lowest energy level, is denoted by n=1. As n increases, the energy levels get closer together and converge to (horizontally) zero at n=, which represents a free electron that is infinitely far from the influence of the nucleus. This unique structure of energy levels is why the hydrogen atom displays specific wavelengths (colors) in its emission spectrum.

Photon Energy Calculation

The energy of a photon released during electronic transitions between energy levels in an atom can be calculated by the difference in energy between the two states. When an electron drops from a higher energy level to a lower one, the energy difference is emitted as a photon of light. The energy (E_{diff}) of such a photon is given by:

\[\begin{equation}E_{diff} = E_{higher} - E_{lower}\end{equation}\]
Using the energy levels formula for hydrogen, we can determine the energy of a specific transition. For instance, the highest possible energy photon corresponds to the transition from the highest occupied energy level to the ground state (n=1). Conceptually, as the difference between the initial and final energy levels increases, the energy of the emitted photon also increases, hence the highest energy photon will come from an electron dropping from the farthest possible level (theoretically, n=, where the energy level is 0 eV) down to n=1, with an energy of -13.6 eV. This calculation reveals that the highest energy photon that can be emitted by an electron in hydrogen is 13.6 eV.

Quantum Transitions

Quantum transitions are the jumps that electrons make between energy levels within an atom. These are governed by quantum mechanics, which allows only certain allowed transitions. An electron transitions to a lower energy level by releasing energy in the form of a photon and to a higher energy level by absorbing a photon of precise energy. This process is responsible for the production of the emission and absorption spectra of elements.
Each transition corresponds to a specific energy change (E_{diff}), which determines the wavelength (\lambda) of the photon through the equation:\[\begin{equation}E_{diff} = \frac{hc}{\lambda}\end{equation}\]Here, 'h' is Planck's constant, and 'c' is the speed of light. The larger the energy change, the higher the frequency and the shorter the wavelength of the emitted light. In a hydrogen atom, the emission spectrum is produced by electrons making transitions between discrete energy levels, commonly from a higher level down to the second level (n=2), resulting in visible spectral lines known as the Balmer series. The highest energy transition, as discussed previously, is from n= to n=1, which does not correspond to visible light but rather to a photon in the ultraviolet range of the electromagnetic spectrum.