An atom in the n=3 state can emit a photon with energies of what values during a single transition?

An atom in the n=3 state can emit a photon with energies of what values during a single transition?

• A. 3 eV
• B. 4 eV or 7 eV only
• C. 7 eV
• D. 4 eV

Answer: (B)

In a hydrogen atom, the energy levels are quantized and can be described by the formula: \[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \] This means for different principal quantum numbers (n), the energy levels can be calculated. For n=3, the energy of the level is: \[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \] To find the possible emitted photon energies during transitions from n=3 to lower levels (n=2 and n=1), we must calculate the energy differences between these states. 1. For the transition from n=3 to n=2: \[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV} \] The energy difference is: \[ \Delta E = E_2 - E_3 = -3.4

In a hydrogen atom, the energy levels are quantized and can be described by the formula:

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

This means for different principal quantum numbers (n), the energy levels can be calculated. For n=3, the energy of the level is:

[ E_3 = -\frac{13.6 , \text{eV}}{3^2} = -\frac{13.6 , \text{eV}}{9} \approx -1.51 , \text{eV} ]

To find the possible emitted photon energies during transitions from n=3 to lower levels (n=2 and n=1), we must calculate the energy differences between these states.

1. For the transition from n=3 to n=2:

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

The energy difference is:

[ \Delta E = E_2 - E_3 = -3.4