Question

According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

• A. n = 5 to n = 3
• B. n = 6 to n = 1
• C. n = 5 to n = 4
• D. n = 6 to n = 5

Answer 1

The Correct Option is D

To determine which transition in the hydrogen atom emits the least energetic photon according to Bohr's theory, we need to understand how energy transitions work in an atom.

According to Bohr's theory, the energy of an electron in a hydrogen atom is quantized and depends on the principal quantum number n. The energy levels are given by the formula: E_n = -\frac{13.6 \, \text{eV}}{n^2} where n is an integer (1, 2, 3, ...).

The energy emitted or absorbed during a transition between two levels is the difference between their energies: \Delta E = E_{\text{initial}} - E_{\text{final}} = \left(-\frac{13.6}{n_{\text{initial}}^2}\right) - \left(-\frac{13.6}{n_{\text{final}}^2}\right)

A transition that gives rise to the least energetic photon will have the smallest difference in energy levels. Let us calculate the energy change for each of the given options:

1. Transition: n = 5 \rightarrow n = 3
   \Delta E = -\frac{13.6}{3^2} + \frac{13.6}{5^2} \approx -1.51 + 0.544 = -0.966 \, \text{eV}
2. Transition: n = 6 \rightarrow n = 1
   \Delta E = -\frac{13.6}{1^2} + \frac{13.6}{6^2} \approx -13.6 + 0.378 = -13.222 \, \text{eV}
3. Transition: n = 5 \rightarrow n = 4
   \Delta E = -\frac{13.6}{4^2} + \frac{13.6}{5^2} \approx -0.85 + 0.544 = -0.306 \, \text{eV}
4. Transition: n = 6 \rightarrow n = 5
   \Delta E = -\frac{13.6}{5^2} + \frac{13.6}{6^2} \approx -0.544 + 0.378 = -0.166 \, \text{eV}

From the calculations, we see that the transition from n = 6 to n = 5 has the smallest energy change, hence resulting in the least energetic photon.

Therefore, the correct answer is: n = 6 to n = 5.