Question

Let \(T_1\) and \(T_2\) be the energy of an electron in the first and second excited states of the hydrogen atom, respectively. According to Bohr’s model of an atom; the ratio T1:T2 is:

• A. \(1:4\)
• B. \(4:1\)
• C. \(4:9\)
• D. \(9:4\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In the Bohr model of the hydrogen atom, the electron revolves around the nucleus in specific quantized orbits.
Each orbit is associated with a specific energy level denoted by the principal quantum number \(n\).
Key Formula or Approach:
The energy of an electron in the \(n^{th}\) orbit of a hydrogen atom is given by:
\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]
This implies that the energy \(E_n\) is inversely proportional to the square of the principal quantum number:
\[ E_n \propto \frac{1}{n^2} \]
Step 2: Detailed Explanation:
1. Ground State: This corresponds to the lowest energy level, \(n = 1\).
2. First Excited State: This corresponds to the second energy level, \(n = 2\).
Therefore, for \(T_1\), \(n_1 = 2\).
\[ T_1 \propto \frac{1}{2^2} = \frac{1}{4} \]
3. Second Excited State: This corresponds to the third energy level, \(n = 3\).
Therefore, for \(T_2\), \(n_2 = 3\).
\[ T_2 \propto \frac{1}{3^2} = \frac{1}{9} \]
4. Calculating the Ratio:
\[ \frac{T_1}{T_2} = \frac{1/n_1^2}{1/n_2^2} = \frac{n_2^2}{n_1^2} \]
\[ \frac{T_1}{T_2} = \frac{3^2}{2^2} = \frac{9}{4} \]
Step 3: Final Answer:
The ratio \(T_1 : T_2\) is \(9 : 4\).