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 Question:
This question asks for the ratio of the orbital velocities of an electron in two different single-electron systems: the fourth orbit ($n_1 = 4$) of a beryllium ion ($\text{Be}^{3+}$) and the second orbit ($n_2 = 2$) of a helium ion ($\text{He}^+$).

Step 2: Key Formula or Approach:
In Bohr's model of the hydrogen-like atom, the velocity ($v_n$) of an electron in the $n$-th orbit is given by the formula:
\[ v_n = v_0 \frac{Z}{n} \] Where:
- $v_0$ is a constant (velocity of electron in the first Bohr orbit of hydrogen, $\approx 2.18 \times 10^6\text{ m/s}$).
- $Z$ is the atomic number of the element.
- $n$ is the principal quantum number (orbit).
Thus, the velocity is directly proportional to $Z/n$:
\[ v_n \propto \frac{Z}{n} \]

Step 3: Detailed Explanation:
Let's calculate the value of $Z/n$ for both systems:

• 1. For Be$^{3+$ (fourth orbit):}
- Beryllium ($\text{Be}$) has an atomic number $Z_1 = 4$.
- The orbit number is $n_1 = 4$.
- The ratio factor is:
\[ \frac{Z_1}{n_1} = \frac{4}{4} = 1 \]
• 2. For He$^+$ (second orbit):
- Helium ($\text{He}$) has an atomic number $Z_2 = 2$.
- The orbit number is $n_2 = 2$.
- The ratio factor is:
\[ \frac{Z_2}{n_2} = \frac{2}{2} = 1 \]
• 3. Calculate the velocity ratio:
- Using our proportionality:
\[ \frac{v_{\text{Be}^{3+}}}{v_{\text{He}^+}} = \frac{Z_1 / n_1}{Z_2 / n_2} = \frac{1}{1} = 1 \]

Step 4: Final Answer:
The ratio of the velocities is $1:1$, which corresponds to option (A).