Question

What is the ratio of the velocity of an electron in the fourth orbit of Be$^{3+}$ to the velocity of the electron in the second orbit of He$^{+}$?

• A. 1:1
• B. 1:2
• C. 3:2
• D. 6:1

Hint:

Bohr's orbital variables are extremely useful for quick calculations:
- Velocity $v \propto Z/n$
- Radius $r \propto n^2/Z$
- Energy $E \propto Z^2/n^2$
For these two systems, both have a $Z/n$ ratio of $1$, which means the electron travels at the exact same speed in both orbits!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Bohr's atomic model gives formulas for the velocity of electrons in hydrogen-like ions.
Key Formula or Approach:
The velocity of an electron in the $n$-th orbit is:
\[ v_{n} = v_{0} \frac{Z}{n} \]
where $v_{0} \approx 2.18 \times 10^{6} \text{ m/s}$, $Z$ is the atomic number, and $n$ is the orbit number.

Step 2: Detailed Explanation:
$\bullet$ For $Be^{3+$ (fourth orbit):}
Atomic number $Z_{1} = 4$ (Beryllium).
Orbit number $n_{1} = 4$.
$v_{Be^{3+}} \propto \frac{4}{4} = 1$.
$\bullet$ For $He^{+$ (second orbit):}
Atomic number $Z_{2} = 2$ (Helium).
Orbit number $n_{2} = 2$.
$v_{He^{+}} \propto \frac{2}{2} = 1$.
$\bullet$ Ratio:
\[ \text{Ratio} = \frac{v_{Be^{3+}}}{v_{He^{+}}} = \frac{1}{1} = 1:1 \]

Step 3: Final Answer:
The ratio is 1:1.
This corresponds to option (A).