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 Question:
The question asks us to calculate the time period of revolution of an electron revolving in the fourth orbit ($n=4$) of a helium ion ($\text{He}^+$, $Z=2$).

Step 2 : Key Formulas and Approach:
The time period of revolution ($T$) is the distance of one orbit divided by the velocity of the electron:
\[ T = \frac{2\pi r}{v} \]
According to Bohr's model:
\[ r_n = a_0 \frac{n^2}{Z} \quad \text{and} \quad v_n = v_0 \frac{Z}{n} \]
Thus, the time period $T_n$ scales as:
\[ T_n \propto \frac{n^3}{Z^2} \]
The standard time period for the ground state of hydrogen ($n=1$, $Z=1$) is:
\[ T_{\text{H}, 1} = \frac{2\pi a_0}{v_0} \approx 1.52 \times 10^{-16}\text{ s} \]

Step 3 : Detailed Explanation:
Let us calculate $T$ for $n = 4$ and $Z = 2$ ($\text{He}^+$):
\[ T = T_{\text{H}, 1} \times \frac{n^3}{Z^2} \]
Plugging in the values:
\[ T = (1.52 \times 10^{-16}\text{ s}) \times \frac{4^3}{2^2} \]
\[ T = (1.52 \times 10^{-16}\text{ s}) \times \frac{64}{4} \]
\[ T = (1.52 \times 10^{-16}\text{ s}) \times 16 \]
\[ T = 2.432 \times 10^{-15}\text{ s} \]
Since $1\text{ femtosecond (fs)} = 10^{-15}\text{ s}$, we have:
\[ T \approx 2.4\text{ femtoseconds} \]

Step 4 : Final Answer:
The time period of revolution of the electron is approximately 2.4 femtoseconds.
This corresponds to Option (A).