Question

An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is:

• A. E = 2U
• B. 2E = 3U
• C. E = U
• D. 2E = U

Answer 1

The Correct Option is D

The electrostatic force between the electron and nucleus is given by:

\[ F = \frac{K(Ze)(e)}{r^2} = \frac{mv^2}{r} \]

Kinetic energy (KE) of the electron is:

\[ \text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \frac{K(Ze)(e)}{r} \]

Potential energy (PE) is given by:

\[ \text{PE} = -\frac{K(Ze)(e)}{r} \]

Total energy (TE) is:

\[ \text{TE} = \text{KE} + \text{PE} = \frac{K(Ze)(e)}{2r} + \left( -\frac{K(Ze)(e)}{r} \right) = -\frac{K(Ze)(e)}{2r} \]

Thus, the relationship between total energy and potential energy is:

\[ 2 \times \text{TE} = \text{PE} \]

Therefore, \( 2E = U \), which corresponds to Option (4).

Answer 2

Step 1: Write expressions for the forces involved
For an electron rotating in a circular orbit around a nucleus of charge \( +Ze \), the electrostatic (Coulomb) force provides the centripetal force. \[ \frac{m v^2}{r} = \frac{1}{4 \pi \varepsilon_0} \frac{Z e^2}{r^2}. \] Thus, \[ m v^2 = \frac{1}{4 \pi \varepsilon_0} \frac{Z e^2}{r}. \]

Step 2: Express kinetic and potential energies
Kinetic energy: \[ K = \frac{1}{2} m v^2 = \frac{1}{8 \pi \varepsilon_0} \frac{Z e^2}{r}. \] Potential energy (electrostatic): \[ U = -\frac{1}{4 \pi \varepsilon_0} \frac{Z e^2}{r}. \] Hence, we see that \[ U = -2K. \]

Step 3: Relation between total energy and potential energy
Total energy: \[ E = K + U = K - 2K = -K. \] Since \(U = -2K\), we get: \[ U = 2E. \]

Step 4: Final conclusion
The correct relation between total energy and potential energy for an electron in a hydrogen-like atom is: \[ \boxed{2E = U.} \]

Final answer

2E = U