Askiitians Tutor Team
Last Activity: 15 Days ago
To understand the scenario you've presented, let's break down the interactions between the photons and the hydrogen atom step by step. The key here is to analyze the energy levels of the hydrogen atom and how they interact with the incoming photons.
Photon and Hydrogen Atom Interaction
When a photon collides with a hydrogen atom, it can either be absorbed or scattered. In your case, we have two distinct photons with energies of 10.2 eV and 15.0 eV, respectively. The ground state energy of a hydrogen atom is -13.6 eV, and the first excited state is -3.4 eV. The energy difference between these two states is 10.2 eV, which is crucial for our analysis.
First Collision: 10.2 eV Photon
When the first photon of 10.2 eV strikes the stationary hydrogen atom, it provides just enough energy to excite the atom from its ground state to the first excited state. This transition can be represented as:
- Ground State (-13.6 eV) + Photon (10.2 eV) = Excited State (-3.4 eV)
After this interaction, the hydrogen atom is now in an excited state. However, this process is inelastic because the photon is absorbed, and the atom gains energy. Importantly, no electron is ejected at this stage.
Second Collision: 15.0 eV Photon
Next, we consider the second photon with an energy of 15.0 eV. The hydrogen atom is currently in the excited state at -3.4 eV. When the 15.0 eV photon collides with the atom, we need to calculate the total energy available:
- Excited State (-3.4 eV) + Photon (15.0 eV) = Total Energy (11.6 eV)
This total energy of 11.6 eV is above the ionization energy of the hydrogen atom (which is 13.6 eV). Therefore, the photon can’t ionize the atom directly. Instead, the atom can absorb some of the energy and eject an electron. The energy of the ejected electron can be calculated as follows:
- Energy of Photon (15.0 eV) - Ionization Energy (13.6 eV) = Energy of Ejected Electron (1.4 eV)
Detection Outcomes
After these interactions, we can summarize what can be detected:
- The first photon (10.2 eV) is absorbed, and thus it cannot be detected afterward.
- The second photon (15.0 eV) is also absorbed, resulting in the ejection of an electron with an energy of 1.4 eV.
In conclusion, the detection setup would identify the ejected electron with an energy of 1.4 eV as a result of the second photon interaction. Therefore, the correct answer is that one photon of 10.2 eV is absorbed, and an electron of energy 1.4 eV is emitted, which can indeed be detected by a suitable detector.
