What energy photon can atoms in the n = 2 state emit spontaneously?

What energy photon can atoms in the n = 2 state emit spontaneously?

• A. 4 eV only
• B. 7 eV only
• C. 3 eV and 4 eV only
• D. 3 eV, 4 eV, and 7 eV

Answer: (A)

To determine what energy photons can be emitted by atoms in the n = 2 state, it is important to analyze the electronic energy levels of hydrogen-like atoms. The energy levels of hydrogen can be calculated using the formula: \[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \] For n = 1 (ground state), the energy is -13.6 eV. For n = 2, the energy is: \[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \] When an electron transitions from a higher energy level to a lower energy level, it emits a photon with energy equal to the difference in energy between those two states. In the case of the n = 2 level: - The next higher energy level is n = 3, which has an energy of: \[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13

To determine what energy photons can be emitted by atoms in the n = 2 state, it is important to analyze the electronic energy levels of hydrogen-like atoms. The energy levels of hydrogen can be calculated using the formula:

[ E_n = -\frac{13.6 , \text{eV}}{n^2} ]

For n = 1 (ground state), the energy is -13.6 eV. For n = 2, the energy is:

[ E_2 = -\frac{13.6 , \text{eV}}{2^2} = -\frac{13.6 , \text{eV}}{4} = -3.4 , \text{eV} ]

When an electron transitions from a higher energy level to a lower energy level, it emits a photon with energy equal to the difference in energy between those two states.

In the case of the n = 2 level:

• The next higher energy level is n = 3, which has an energy of:

[ E_3 = -\frac{13.6 , \text{eV}}{3^2} = -\frac{13