Question

What is the time period of revolution of an electron in the fourth Bohr orbit of $\text{He}^+$?
(Bohr radius = 52.9 picometers, mass of an electron = $9.11 \times 10^{-31}\text{ kg}$, Planck’s constant = $6.626 \times 10^{-34}\text{ Js}$)}

• A. 2.4 femtoseconds
• B. 4.8 femtoseconds
• C. 24 femtoseconds
• D. 0.24 femtoseconds

Hint:

Remember the scaling relation:
\[ T \propto \frac{n^3}{Z^2} \]
For $n=4$ and $Z=2$, the factor is $\frac{64}{4} = 16$.
Multiplying the base ground-state hydrogen time period ($1.5 \times 10^{-16}\text{ s}$) by 16 gives $2.4 \times 10^{-15}\text{ s}$ directly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Bohr's model of the atom, the time period of revolution (\( T \)) of an electron in a given orbit is the total time required to complete one full circular path around the nucleus. It depends on the radius of the orbit (\( r \)) and the orbital velocity (\( v \)) of the electron.
Step 2: Key Formula or Approach:
The standard formula for the time period is:
\[ T = \frac{2\pi r}{v} \]
In Bohr's model, the relationships for radius and velocity as functions of the principal quantum number (\( n \)) and atomic number (\( Z \)) are:
- \( r_n = r_0 \frac{n^2}{Z} \), where \( r_0 = 52.9 \text{ pm} = 0.529 \text{ \AA} \)
- \( v_n = v_0 \frac{Z}{n} \), where \( v_0 \approx 2.18 \times 10^6 \text{ m/s} \)
Substituting these into the time period expression yields:
\[ T \propto \frac{n^2/Z}{Z/n} \implies T \propto \frac{n^3}{Z^2} \]
The standard time period for the ground state of hydrogen (\( n=1, Z=1 \)) is:
\[ T_1 = \frac{2\pi (0.529 \times 10^{-10} \text{ m})}{2.18 \times 10^6 \text{ m/s}} \approx 1.52 \times 10^{-16} \text{ s} \]
Thus, the general equation for any hydrogen-like species is:
\[ T = T_1 \left( \frac{n^3}{Z^2} \right) \]
Step 3: Detailed Explanation:
Given parameters from the problem:
- Species: \( \text{He}^+ \implies Z = 2 \)
- Orbit number: \( n = 4 \)
Let us compute the ratio \( \frac{n^3}{Z^2} \):
\[ \frac{n^3}{Z^2} = \frac{4^3}{2^2} = \frac{64}{4} = 16 \]
Now, calculate the absolute value of the time period \( T \):
\[ T = 1.52 \times 10^{-16} \text{ s} \times 16 \]
\[ T = 24.32 \times 10^{-16} \text{ s} \]
Converting this to standard scientific notation:
\[ T = 2.432 \times 10^{-15} \text{ s} \]
Since \( 1 \text{ femtosecond (fs)} = 10^{-15} \text{ s} \):
\[ T \approx 2.4 \text{ fs} \]
Step 4: Final Answer:
The time period of revolution of the electron is 2.4 femtoseconds.