Question

The velocity of an electron in the seventh orbit of hydrogen-like atom is 3.6 × 10⁶ m/s. Find the velocity of the electron in the 3^rd orbit.

• A. 4.2 × 10⁶ m/s
• B. 8.4 × 10⁶ m/s
• C. 2.1 × 10⁶ m/s
• D. 3.6 × 10⁶ m/s

Answer 1

The Correct Option is B

Given:

• The velocity of the electron in the 7th orbit is \(3.6 \times 10^6 \, \text{m/s}\).
• The velocity of the electron in the 3rd orbit is unknown.

Formula:

The velocity of an electron in a hydrogen-like atom is inversely proportional to the principal quantum number \(n\), given by the relation:

\[ \frac{v_1}{v_2} = \frac{n_2}{n_1} \]

Where:

• \(v_1\) is the velocity of the electron in the 7th orbit \( (v_7) \).
• \(v_2\) is the velocity of the electron in the 3rd orbit \( (v_3) \).
• \(n_1\) is the principal quantum number for the 7th orbit.
• \(n_2\) is the principal quantum number for the 3rd orbit.

Solution:

Using the formula:

\[ \frac{v_7}{v_3} = \frac{3}{7} \]

Substitute the given values:

\[ \frac{3.6 \times 10^6}{v_3} = \frac{3}{7} \]

Now, solve for \(v_3\):

\[ v_3 = \frac{3.6 \times 10^6 \times 7}{3} = 8.4 \times 10^6 \, \text{m/s} \]

Final Answer: The velocity of the electron in the 3rd orbit is \(8.4 \times 10^6 \, \text{m/s}\).