(a) Find the highest possible energy for a photon emitted as the electron jumps between two adjacent energy levels in the Bohr hydrogen atom. (b) Which energy levels are involved?

Short Answer

Expert verified
The maximum possible energy for a photon emitted during a jump between two adjacent energy levels in a Bohr hydrogen atom is 10.2 eV. The adjacent energy levels involved are 2 (initial) and 1 (final, ground state).

Step by step solution

01

Understanding the Formula

The formula for the energy difference, which gives us the energy of the emitted photon, when an electron jumps from energy level \(n_{i}\) to \(n_{f}\) in a hydrogen atom is given by the Rydberg formula: \[ E = 13.6 eV * \left( \frac{1}{{n_{f}^{2}}} - \frac{1}{{n_{i}^{2}}} \right) \]Here, \(13.6 eV\) is the Rydberg constant for hydrogen. \(n_{i}\) and \(n_{f}\) represent the initial and final energy levels. As the energy is emitted during a jump from a higher level to a lower one, \(n_{i}\) should be greater than \(n_{f}\). We are looking for the highest possible energy, hence \(n_{f}\) should be the lowest possible.
02

Implementing the Energy Formula

Considering two adjacent energy levels, the level difference \(n_{i} - n_{f}\) should be 1. As we are considering the lowest possible value for \(n_{f}\) to be 1 (the ground state), \(n_{i}\) becomes 2. Substituting these values into the Rydberg formula we get:\[ E = 13.6 eV * \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
03

Calculating the Maximum Energy

Calculating the above expression:\[ E = 13.6 eV * \left(1 - \frac{1}{4}\right) \]\[ E = 13.6 eV * \frac{3}{4} = 10.2 eV \]This is the maximum possible energy for a photon emitted as the electron jumps between two adjacent energy levels in a Bohr hydrogen atom.
04

Identifying the Energy Levels Involved

The adjacent energy levels involved are the initial energy level 2 and the final energy level 1, which is the ground state. These are the levels that give the maximum energy of the emitted photon as per the Bohr hydrogen atom model.

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