To determine the energy associated with the $n = 7$ energy state of a hydrogen atom, we use the formula for the energy levels of hydrogen:

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

Convert this energy to joules using the conversion factor $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$.

Step 1: Plug $n = 7$ into the formula

$E_7 = -\frac{13.6}{7^2} \, \text{eV}$ $E_7 = -\frac{13.6}{49} \, \text{eV}$ $E_7 = -0.2776 \, \text{eV}$

Step 2: Convert to joules

$E_7 = -0.2776 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV}$ $E_7 = -4.45 \times 10^{-20} \, \text{J}$

Final Answer:

The energy associated with the $n = 7$ energy state is: $E_7 = -4.45 \times 10^{-20} \, \text{J}$

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Related Questions:

1. What is the energy in joules for the $n = 1$ state of a hydrogen atom?
2. How do energy level differences relate to photon emission or absorption?
3. Why is the energy negative in this context?
4. What is the relationship between the energy level formula and the Bohr model?
5. How do quantum numbers affect other properties like angular momentum?

Tip:

Always check units and conversion factors when switching between energy in eV and joules. This ensures precision in physics calculations.