Question

An electron makes a transition from the \( n = 2 \) level to the \( n = 1 \) level in the Bohr model of a hydrogen atom. Its period of revolution:

• A. Increases by 87.5\%
• B. Decreases by 87.5\%
• C. Increases by 43.75\%
• D. Decreases by 43.75\%

Hint:

Remember that in the Bohr model, the period of revolution of the electron around the nucleus is directly proportional to the cube of the principal quantum number \( n \).

Answer 1

The Correct Option is B

To determine how the period of revolution of an electron changes when it transitions from the \( n = 2 \) level to the \( n = 1 \) level in the Bohr model of a hydrogen atom, we can follow these steps: 1. Bohr Model Basics
In the Bohr model, the period of revolution \( T \) of an electron in the \( n \)-th energy level is given by:
\[ T_n \propto n^3 \] This means that the period of revolution is proportional to the cube of the principal quantum number \( n \).
2. Initial and Final Periods
\begin{itemize} \item For \( n = 2 \): \[ T_2 \propto 2^3 = 8 \] \item For \( n = 1 \): \[ T_1 \propto 1^3 = 1 \] \end{itemize} 3. Change in Period
The change in the period of revolution when the electron transitions from \( n = 2 \) to \( n = 1 \) is:
\[ \Delta T = T_1 - T_2 = 1 - 8 = -7 \] The negative sign indicates a decrease in the period.
4. Percentage Change
The percentage change in the period is calculated as:
\[ Percentage Change} = \left( \frac{\Delta T}{T_2} \right) \times 100\% = \left( \frac{-7}{8} \right) \times 100\% = -87.5\% \] This means the period of revolution decreases by 87.5\%. Therefore, the correct answer is: \[ \boxed{(B) decreases by 87·5\%}} \]