Question

The speed of the electron in a hydrogen atom in the \( n = 3 \) level is:

• A. \( 6.2 \times 10^5 \) ms\(^{-1}\)
• B. \( 3.7 \times 10^5 \) ms\(^{-1}\)
• C. \( 7.3 \times 10^5 \) ms\(^{-1}\)
• D. \( 1.6 \times 10^5 \) ms\(^{-1}\)

Hint:

In quantum mechanics, the orbital speed of an electron decreases with increasing principal quantum number \( n \) because the electron is further from the nucleus and experiences less electrostatic force.

Answer 1

The Correct Option is C

Step 1: The speed \( v \) of an electron in the nth orbit of a hydrogen atom can be calculated using the formula: \[ v = \frac{e^2}{2\epsilon_0 h} \cdot \frac{1}{n} \] where: - \( e \) is the elementary charge (\(1.6 \times 10^{-19} \, C\)) - \( \epsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \, {C}^2 \, {N}^{-1} \, {m}^{-2}\)) - \( h \) is Planck's constant (\(6.63 \times 10^{-34} \, {J} \cdot {s}\)) - \( n \) is the principal quantum number
Step 2: For \( n = 3 \), substitute the known values into the formula: \[ v = \frac{(1.6 \times 10^{-19})^2}{2 \times 8.85 \times 10^{-12} \times 6.63 \times 10^{-34}} \cdot \frac{1}{3} \] Simplifying this expression yields: \[ v \approx 7.3 \times 10^5 \, {ms}^{-1} \]