Question

According to Bohr's first postulate the kinetic energy \(\frac{1}{2} m v^2\) of the electron in C.G.S. system is ( \(\mathrm{m}=\) mass of electron, v is its velocity, r is the radius of the stationary orbit around the nucleus with charge Ze )

• A. \(\frac{\mathrm{Ze}}{\mathrm{r}^{2}}\)
• B. \(\frac{\mathrm{Ze}^{2}}{\mathrm{r}}\)
• C. \(\frac{\mathrm{Ze}^{2}}{2\mathrm{r}}\)
• D. \(\frac{\mathrm{Ze}^{2}}{2\mathrm{r}^{2}}\)

Hint:

The electrostatic force in CGS units gives the centripetal force.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Bohr's first postulate: the electrostatic attraction provides the centripetal force. In C.G.S., Coulomb force is \(\dfrac{Ze\cdot e}{r^{2}}\).

Step 2: Equation
\(\dfrac{mv^{2}}{r}=\dfrac{Ze^{2}}{r^{2}}\)
so \(mv^{2}=\dfrac{Ze^{2}}{r}\).

Step 3: Calculation
Kinetic energy \(=\dfrac{1}{2}mv^{2}=\dfrac{Ze^{2}}{2r}\).

Step 4: Compare the options
This is option (C).

Final Answer:
The kinetic energy is \(\dfrac{Ze^{2}}{2r}\). \[ \boxed{\text{(C) }\dfrac{\mathrm{Ze}^{2}}{2\mathrm{r}}} \]