An atom absorbs a yellow photon in the transition from its 2s to its 3s state (see figure on the left, not drawn to scale). What color photon might it emit in a transition from the 3s to the 2p state?
Your figure is certainly not "drawn to scale." It isn't drawn at all.
For hydrogen, the energy of levels 2p and 2s are the same. Therefore the emitted color would be yellow, the same as the absorbed color.
Your missing figure may depict a different situation.
To determine the color of the photon that might be emitted in the transition from the 3s to the 2p state, we need to understand the energy levels and transitions within an atom.
In an atom, electrons occupy distinct energy levels or shells around the nucleus. These energy levels are labeled with quantum numbers, such as 1s, 2s, 2p, 3s, etc. When an electron transitions between different energy levels, it can either absorb or emit a photon. The energy of the photon is directly related to the energy difference between the two energy levels involved in the transition.
In this case, the atom absorbs a yellow photon during the transition from the 2s to the 3s state. This tells us that the energy difference between the 2s and 3s states corresponds to the energy of a yellow photon.
To determine the color of the photon that might be emitted in the transition from the 3s to the 2p state, we need to find the energy difference between these two states. The energy difference will determine the color of the photon emitted.
The energies of different energy levels in an atom can be calculated using the Rydberg equation:
1/λ = RZ²(1/n₁² - 1/n₂²)
Where:
- λ is the wavelength of the photon
- R is the Rydberg constant (approximately 1.097 × 10^7 m⁻¹)
- Z is the atomic number (number of protons in the nucleus)
- n₁ and n₂ are the principal quantum numbers of the initial and final energy levels, respectively
For the transition from the 2s to the 3s state, we know that a yellow photon is absorbed. Since the color yellow corresponds to a specific wavelength (around 570-590 nanometers), we can use the wavelength to find the energy difference between the 2s and 3s states.
1/λ = RZ²(1/n₁² - 1/n₂²)
By substituting the known values, including the wavelength corresponding to yellow light, you can solve for the initial and final quantum numbers (n₁ and n₂).
Once you have determined the values of n₁ and n₂ for the 3s to 2p transition, you can use the Rydberg equation again to find the corresponding wavelength or color of the emitted photon.