Question

According to Bohr's model of the hydrogen atom, the ratio of the kinetic energy to the total energy of an electron in \(3^{\text{rd}}\) excited state is?

• A. \( 1 : 1 \)
• B. \( 1 : -1 \)
• C. \( -1 : 1 \)
• D. \( 1 : 2 \)

Hint:

Notice that the question asks for a structural ratio matching specific energy types. Because the proportional relation \(\text{KE} = -\text{TE}\) is a universal constant rule valid for any allowed Bohr orbit, the ratio will always equal \(1 : -1\) regardless of whether the electron sits in the ground state, the \(3^{\text{rd}}\) excited state, or the \(100^{\text{th}}\) orbit! The state value is extra distracting information.

Answer 1

The Correct Option is B

Concept: In Bohr's atomic model, an electron revolving around a nucleus of atomic number \(Z\) in an orbit of principal quantum number \(n\) possesses kinetic energy (\(\text{KE}\)), potential energy (\(\text{PE}\)), and total energy (\(\text{TE}\)). These energies are linked by a fundamental proportional identity: \[ \text{KE} = -\text{TE} = -\frac{\text{PE}}{2} \] Where:
• Kinetic energy is always positive (\(\text{KE} > 0\)) as it depends on the square of the orbital velocity.
• Total energy is negative (\(\text{TE} < 0\)), signifying that the electron is bound within the attractive electrostatic field of the nucleus.

Step 1: Analyzing the principal quantum number for the given state.
The problem specifies that the electron is in the \(3^{\text{rd}}\) excited state. In atomic energy levels:
• Ground state matches \(n = 1\)
• \(1^{\text{st}}\) excited state matches \(n = 2\)
• \(2^{\text{nd}}\) excited state matches \(n = 3\)
• \(3^{\text{th}}\) excited state matches \(n = 4\) Thus, the principal quantum number for this electron configuration is \(n = 4\).

Step 2: Evaluating the energy ratio.
The standard formula for the total energy of an electron in the \(n^{\text{th}}\) orbit of a hydrogen atom (\(Z=1\)) is: \[ \text{TE}_n = -\frac{13.6}{n^2} \text{ eV} \] For the \(3^{\text{rd}}\) excited state (\(n=4\)): \[ \text{TE}_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \text{ eV} \] Since kinetic energy is equal in magnitude but opposite in sign to the total energy (\(\text{KE} = -\text{TE}\)): \[ \text{KE}_4 = -(-0.85 \text{ eV}) = +0.85 \text{ eV} \] Now, computing the required ratio of kinetic energy to total energy: \[ \text{Ratio} = \frac{\text{KE}_4}{\text{TE}_4} = \frac{0.85 \text{ eV}}{-0.85 \text{ eV}} = \frac{1}{-1} = 1 : -1 \]