Question:medium

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?

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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.
Updated On: May 29, 2026
  • \( 1 : 1 \)
  • \( 1 : -1 \)
  • \( -1 : 1 \)
  • \( 1 : 2 \)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
The question belongs to the topic of atomic structure in modern physics, specifically focusing on Bohr's model of the hydrogen atom.
We are asked to determine the ratio of the kinetic energy to the total energy of an electron in its third excited state.
By understanding the relationship between different forms of energy in an orbit, we can solve this problem directly.
Step 2: Key Formulas and Approach:
In Bohr's model, the kinetic energy ($\text{KE}$), potential energy ($\text{PE}$), and total energy ($\text{TE}$) of an electron are related by a standard proportionality.
The fundamental relationship between these quantities in any stable orbit is:
\[ \text{KE} = -\text{TE} = -\frac{\text{PE}}{2} \] Our approach will be to apply this universal relation to express the ratio of kinetic energy to total energy.
Step 3: Detailed Explanation:

Analyze the Energy Expressions: The kinetic energy of an electron revolving in the $n^{\text{th}}$ orbit is always a positive quantity, given by $\text{KE} = \frac{13.6 \cdot Z^2}{n^2}\text{ eV}$.

Analyze the Total Energy: The total energy of the electron in the same orbit is negative, representing a bound state, and is given by $\text{TE} = -\frac{13.6 \cdot Z^2}{n^2}\text{ eV}$.

Determine the principal quantum number ($n$): The problem mentions the $3^{\text{rd}}$ excited state.

The ground state corresponds to $n = 1$.

The first excited state corresponds to $n = 2$.

The second excited state corresponds to $n = 3$.

The third excited state corresponds to $n = 4$.

Calculate the Energies: For $n = 4$ and $Z = 1$ (hydrogen atom):
\[ \text{TE}_4 = -\frac{13.6}{4^2} = -0.85\text{ eV} \] \[ \text{KE}_4 = -(\text{TE}_4) = -(-0.85\text{ eV}) = +0.85\text{ eV} \]
Compute the Ratio: Now we find the ratio of $\text{KE}$ to $\text{TE}$:
\[ \frac{\text{KE}}{\text{TE}} = \frac{0.85\text{ eV}}{-0.85\text{ eV}} = \frac{1}{-1} \]
This ratio simplifies to $1 : -1$, demonstrating that the specific energy state does not affect the proportional relationship.

Step 4: Final Answer:
The ratio of kinetic energy to total energy is $1 : -1$, which corresponds to Option (B).
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