To tackle this problem, we need to analyze the behavior of hydrogen-like atoms in terms of their energy levels and the transitions that occur when they absorb and emit photons. Let's break down the question step by step.
Identifying the Principal Quantum Number of Level B
In hydrogen-like atoms, the energy levels are given by the formula:
E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2}
where Z is the atomic number (which is 1 for hydrogen), and n is the principal quantum number. The energy difference between two levels can be calculated using:
ΔE = E_n - E_m = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} + \frac{Z^2 \cdot 13.6 \, \text{eV}}{m^2}
Given that the photon energy absorbed is 2.7 eV, we can set this equal to the energy difference between the ground state (n=1) and the excited state (n=B):
2.7 \, \text{eV} = -\frac{13.6 \, \text{eV}}{1^2} + \frac{13.6 \, \text{eV}}{B^2}
Rearranging gives:
\frac{13.6 \, \text{eV}}{B^2} = 2.7 + 13.6
\frac{13.6}{B^2} = 16.3
Solving for B:
B^2 = \frac{13.6}{16.3} \implies B \approx 1.01
Since B must be a whole number, we can conclude that the principal quantum number of the initially excited level B is likely 2.
Determining the Ionization Energy
The ionization energy is the energy required to remove an electron from the ground state (n=1) to infinity (n=∞). For hydrogen-like atoms, this is simply:
Ionization Energy = -E_1 = 13.6 \, \text{eV}
Thus, the ionization energy for the gas atoms is 13.6 eV.
Calculating Maximum and Minimum Energy of Emitted Photons
When the atoms transition from the excited state (B) back to the ground state (A), they can emit photons of various energies. The emitted photon energies will depend on the transitions between the energy levels. The maximum energy photon corresponds to the transition from the highest excited state to the ground state, while the minimum energy photon corresponds to a transition to the next lower energy level.
The maximum energy photon emitted when transitioning from level B (2) to level A (1) is:
E_{max} = E_B - E_A = -\frac{13.6}{1^2} + \frac{13.6}{2^2} = -13.6 + 3.4 = 10.2 \, \text{eV}
The minimum energy photon emitted corresponds to a transition from level B to the next lower level (which we can assume to be level 1, as there are no other levels mentioned). Therefore, the minimum energy photon is:
E_{min} = E_B - E_{next} = E_B - E_A = 2.7 \, \text{eV} - 0 = 2.7 \, \text{eV}
Summary of Results
- Principal Quantum Number of Level B: 2
- Ionization Energy: 13.6 eV
- Maximum Energy of Emitted Photons: 10.2 eV
- Minimum Energy of Emitted Photons: 2.7 eV
This analysis illustrates how energy levels in hydrogen-like atoms dictate the behavior of photons during transitions. Understanding these principles is crucial for grasping atomic physics and quantum mechanics.
