Question

Calculate the angular velocity of an electron in a hydrogen atom in the second orbit ($n = 2$).

• A. \(5.15 \times 10^{15} \, \text{rad/s}\)
• B. \(7.15 \times 10^{15} \, \text{rad/s}\)
• C. \(8.15 \times 10^{15} \, \text{rad/s}\)
• D. \(12.25 \times 10^{15} \, \text{rad/s}\)

Hint:

N/A

Answer 1

The Correct Option is A

• The speed of an electron in the \(n^{\text{th}}\) orbit of hydrogen is given by: \[ v = 2.188 \times 10^6 \times \frac{Z}{n} \text{ m/s} \]
• Angular velocity is given by: \[ \omega = \frac{v_n}{r_n} \]
• Using the expression: \[ \omega = \frac{2.18}{5.29} \times \frac{Z^2}{n^3} \times 10^{17} \]
• For hydrogen, \(Z = 1\), and \(n = 2\), so: \[ \omega = \frac{2.18}{5.29} \times \frac{1^2}{2^3} \times 10^{17} \]
• Simplifying: \[ \omega = 5.15 \times 10^{15} \text{ rad/s} \]