Question

A hydrogen atom in an \(n=2\) state absorbs a photon. (a) What should be the photon wavelength to cause the electron to jump to an \(n=4\) state? (b) What wave length photons might be emitted by the atom following this absorption?

Short answer

The wavelength of the absorbed photon should be calculated from the Rydberg formula. Likewise, the possible emission wavelengths following the absorption can be calculated from the same formula, taking into account all possible transitions starting from \(n=4\) state to lower energy states.

Step-by-step solution

Calculate the Absorption Wavelength

Use the Rydberg formula to calculate the wavelength of the photon absorbed to cause the electron to jump from \(n=2\) to \(n=4\). Plug \(n_1 = 2\), \(n_2 = 4\) and \(R = 1.097373 \times 10^7 m^{-1}\) into the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Solve it for wavelength \(\lambda\).

Calculate the Emission Wavelengths

For the wavelengths of photon that might be emitted following this absorption: (i) For the jump from \(n=4\) to \(n=3\), plug \(n_1 = 4\), \(n_2 = 3\) and \(R = 1.097373 \times 10^7 m^{-1}\) into the Rydberg formula and solve it for \(\lambda\). (ii) For the jump from \(n=4\) to \(n=2\), plug \(n_1 = 4\), \(n_2 = 2\) and \(R = 1.097373 \times 10^7 m^{-1}\) into the Rydberg formula and solve it for \(\lambda\). (iii) For the jump from \(n=4\) to \(n=1\), plug \(n_1 = 4\), \(n_2 = 1\) and \(R = 1.097373 \times 10^7 m^{-1}\) into the Rydberg formula and solve it for \(\lambda\). All three possible emission wavelengths should be calculated.

Key concepts

Hydrogen Atom Transitions

Understanding hydrogen atom transitions is key to grasping much of atomic physics. Transitions occur when an electron in a hydrogen atom moves between different energy levels, or "shells." Each energy level is assigned a quantum number, denoted by \( n \). When an electron jumps from a lower energy level to a higher one, it absorbs energy in the form of a photon. Conversely, when it falls from a higher energy level to a lower one, it emits a photon.
In the given problem, the transition in question is from \( n=2 \) to \( n=4 \). This move requires energy to be absorbed, typically from a photon, to excite the electron to the higher energy state. When the electron returns to a lower energy state, it releases energy, often as light. These changes in energy levels are discrete, meaning they happen in specific steps, and each transition corresponds to a particular energy difference.

Photon Absorption

Photon absorption by an atom occurs when an incident photon has just the right amount of energy to move an electron from a lower to a higher energy level. Photon energy depends on its wavelength or frequency; thus, only photons with certain wavelengths will be absorbed.

• Energy of a photon can be calculated using the formula: \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant and \( c \) is the speed of light.
• The wavelength \( \lambda \) is what we tried to calculate in part (a) of the problem.
• This specific wavelength causes the electron in the hydrogen atom to transition from the \( n=2 \) level to the \( n=4 \) level.

The correct wavelength needed for this absorption ensures that the photon's energy matches the energy difference between the two quantum levels involved.

Emission Wavelengths

Emission wavelengths refer to the specific wavelengths of light released when an electron falls from a higher to a lower energy level in an atom. After absorption, an electron temporarily resides in a higher energy state. It will eventually return to a lower energy level, and during this process, it emits a photon with a wavelength corresponding to the energy difference between the levels.

• For instance, part (b) of the problem involves calculating the emission wavelengths when the electron eventually falls from \( n=4 \) to lower levels such as \( n=3 \), \( n=2 \), and \( n=1 \).
• Each of these drops results in a specific photon emission, characterized by its unique wavelength according to the Rydberg formula.

These emitted photons can be detected and are responsible for the spectral lines observed in atomic spectra, revealing much about the atomic structure and transitions.