Question

The energy of electron in an excited hydrogen atom is -3.4 eV. Determine the angular momentum of this electron.

Hint:

It is useful to memorize the energies of the first few levels of the hydrogen atom: \(E_1 = -13.6\) eV, \(E_2 = -3.4\) eV, \(E_3 = -1.51\) eV. Recognizing that -3.4 eV corresponds to the n=2 level can save you time on calculations.

Answer 1

Step 1: Understanding the Concept:
This problem combines two key principles of the Bohr model for the hydrogen atom: the formula for quantized energy levels and the postulate for quantized angular momentum. First, we must identify the principal quantum number (\(n\)) of the electron's orbit from its given energy. Then, we can use this value of \(n\) to calculate the angular momentum.

Step 2: Key Formula or Approach:
1. The energy (\(E_n\)) of an electron in the \(n^{th}\) orbit of a hydrogen atom is given by: \[ E_n = -\frac{13.6}{n^2} \text{ eV} \] 2. The angular momentum (\(L_n\)) of an electron in the \(n^{th}\) orbit is quantized according to Bohr's second postulate: \[ L_n = n \frac{h}{2\pi} \] where \(h\) is Planck's constant (\(h \approx 6.626 \times 10^{-34}\) J·s).

Step 3: Detailed Explanation:
Part 1: Find the Principal Quantum Number (n)
We are given the energy of the electron, \(E_n = -3.4\) eV. We can use the energy formula to find \(n\): \[ -3.4 = -\frac{13.6}{n^2} \] \[ n^2 = \frac{13.6}{3.4} = 4 \] Taking the square root, we get: \[ n = 2 \] This means the electron is in the first excited state (\(n=2\)).
Part 2: Calculate the Angular Momentum (L)
Now that we have \(n=2\), we can calculate the angular momentum using Bohr's quantization rule: \[ L_2 = 2 \times \frac{h}{2\pi} = \frac{h}{\pi} \] Substituting the value of Planck's constant: \[ L_2 = \frac{6.626 \times 10^{-34} \text{ J·s}}{3.14159} \] \[ L_2 \approx 2.11 \times 10^{-34} \text{ J·s} \]

Step 4: Final Answer:
The angular momentum of the electron is approximately \(2.11 \times 10^{-34}\) J·s.