Which principal quantum number, n, corresponds to the orbit with the highest energy in a hydrogen atom?

• A. n = 2
• B. n = 1
• C. n = 3
• D. n = 4

Answer: (D)

Verified step by step guidance

1. Recall that in a hydrogen atom, the energy of an electron in an orbit is given by the formula \(E_n = -\frac{13.6\ \text{eV}}{n^2}\), where \(n\) is the principal quantum number.
2. Understand that the energy levels are quantized and negative, meaning that as \(n\) increases, the energy becomes less negative and thus higher (closer to zero).
3. Compare the energy values for the given principal quantum numbers by substituting \(n = 1, 2, 3,\) and \$4\( into the formula \)E_n = -\frac{13.6}{n^2}$ to see which has the highest energy.
4. Recognize that the orbit with the highest energy corresponds to the largest value of \(n\) because the energy approaches zero from below as \(n\) increases.
5. Conclude that among the options, \(n = 4\) corresponds to the orbit with the highest energy since it has the largest principal quantum number.