Question

The electrostatic potential energy of the electron in an orbit of hydrogen is \( -6.8 \) eV. The speed of the electron in this orbit is (C is the speed of light in vacuum):

• A. \( \frac{C}{137} \)
• B. \( \frac{C}{274} \)
• C. \( \frac{2C}{137} \)
• D. \( \frac{3C}{137} \)

Hint:

The speed of an electron in a hydrogen atom follows the relation \( v = \frac{C}{\alpha n} \), where \( \alpha \approx \frac{1}{137} \) and \( n \) is the principal quantum number.

Answer 1

The Correct Option is B

Step 1: Use the Relationship Between Energy and Potential Energy The electrostatic potential energy is related to the total energy by: \[ U = 2E \] For hydrogen-like atoms: \[ E_n = -\frac{13.6}{n^2} { eV} \] Since given \( U = -6.8 \) eV: \[ 2E_n = -6.8 \] \[ E_n = -3.4 { eV} \] Step 2: Solve for \( n \) \[ -\frac{13.6}{n^2} = -3.4 \] \[ n^2 = \frac{13.6}{3.4} = 4 \] \[ n = 2 \] Step 3: Compute Electron Speed The speed of an electron in an orbit is given by: \[ v = \frac{C}{\alpha n} = \frac{C}{137 \times 2} \] \[ v = \frac{C}{274} \] Thus, the correct answer is \( \frac{C}{274} \).