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 the change in an electron's period of revolution when transitioning from the \( n = 2 \) to the \( n = 1 \) level in the Bohr model of a hydrogen atom, we employ the following methodology:1. Bohr Model Fundamentals
The Bohr model establishes that the period of revolution \( T \) for an electron in the \( n \)-th energy level is proportional to the cube of the principal quantum number \( n \):
\[T_n \propto n^3\]2. Calculating Initial and Final Periods
\begin{itemize} \item For the \( n = 2 \) level: \[ T_2 \propto 2^3 = 8 \] \item For the \( n = 1 \) level: \[ T_1 \propto 1^3 = 1 \]\end{itemize}3. Quantifying the Period Change
The absolute change in the period of revolution during the transition from \( n = 2 \) to \( n = 1 \) is calculated as:\[\Delta T = T_1 - T_2 = 1 - 8 = -7\]The negative sign signifies a reduction in the period.
4. Determining Percentage Change
The percentage change in the period is computed using the formula:\[Percentage Change} = \left( \frac{\Delta T}{T_2} \right) \times 100\% = \left( \frac{-7}{8} \right) \times 100\% = -87.5\%\]This indicates an 87.5\% decrease in the period of revolution.
Consequently, the validated outcome is:\[\boxed{(B) decreases by 87·5\%}}\]