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: \[ E_n = -\frac{13.6}{n^2} \text{ eV} \] Since given \( U = -6.8 \) eV: \[ 2E_n = -6.8 \] \[ E_n = -3.4 \text{ eV} \] 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 \( \frac{C}{274} \). \bigskip

Answer 2

Given:
- Electrostatic potential energy \( U = -6.8 \) eV for an electron in a hydrogen atom orbit.
- \( C \) = speed of light in vacuum.

We need to find the speed \( v \) of the electron in this orbit.

Step 1: Relation between electrostatic potential energy and total energy in hydrogen atom:
In the Bohr model,
\[ U = 2E \] where \( U \) is the electrostatic potential energy and \( E \) is the total energy of the electron.
Given \( U = -6.8 \) eV,
\[ E = \frac{U}{2} = \frac{-6.8}{2} = -3.4 \, \text{eV} \]

Step 2: Total energy in the Bohr model is:
\[ E = - \frac{13.6}{n^2} \, \text{eV} \] Equating and solving for \( n \):
\[ -3.4 = - \frac{13.6}{n^2} \Rightarrow n^2 = \frac{13.6}{3.4} = 4 \Rightarrow n = 2 \]

Step 3: Velocity of the electron in the \( n^{th} \) orbit is given by:
\[ v_n = \frac{Z e^2}{2 \varepsilon_0 h n} \] But more conveniently, velocity can be expressed relative to speed of light as:
\[ v_n = \frac{C}{137} \times \frac{Z}{n} \] For hydrogen, \( Z = 1 \), so:
\[ v_n = \frac{C}{137 \times n} \]

Step 4: Substitute \( n = 2 \):
\[ v = \frac{C}{137 \times 2} = \frac{C}{274} \]

Therefore, the speed of the electron in this orbit is:
\[ \boxed{\frac{C}{274}} \]