For a hydrogen atom, what is the energy (in electron volts, eV) of an electron in the n=4 excited state?

• A. -1.51 eV
• B. -0.85 eV
• C. -3.40 eV
• D. -13.6 eV

Answer: (B)

Verified step by step guidance

1. Recall that the energy levels of a hydrogen atom are given by the formula: \(E_n = -13.6 \times \frac{1}{n^2}\) eV, where \(n\) is the principal quantum number representing the energy level.
2. Identify the given energy level \(n=4\) as the excited state for which we want to find the energy of the electron.
3. Substitute \(n=4\) into the energy formula: \(E_4 = -13.6 \times \frac{1}{4^2}\) eV.
4. Calculate the denominator \$4^2 = 16\(, so the expression becomes \)E_4 = -13.6 \times \frac{1}{16}$ eV.
5. Multiply \(-13.6\) by \(\frac{1}{16}\) to find the energy of the electron in the \(n=4\) state (do not compute the final value here, just set up the expression).