Q148 / 150 · MHT CET 2025 (27 Apr Shift 2) · Physics

If ' m ' is the mass of electron, ' V ' is its velocity, ' r ' is the radius of stationary circular orbit around a nucleus with charge 'Ze' then from Bohr's first postulate the kinetic energy of the electron is $\left(\mathrm{K}=1 / 4 \pi \epsilon_0\right)$

  • Correct answer
    A
    $\frac{\mathrm{Ze}^2}{2 \mathrm{r}} \mathrm{K}$
  • B
    $\frac{\mathrm{Ze}^2}{2 \mathrm{r}^2} \mathrm{~K}$
  • C
    $\frac{\mathrm{Ze}^2}{\mathrm{r}} \mathrm{K}$
  • D
    $\frac{\mathrm{Ze}}{\mathrm{r}^2} \mathrm{~K}$

Solution

The kinetic energy of an electron in a stationary circular orbit is derived from Bohr's first postulate, where the centripetal force equals the electrostatic attraction.

The electrostatic force between nucleus charge $Ze$ and electron charge $e$ is $F_e = K\frac{Ze^2}{r^2}$ with $K = \frac{1}{4\pi\epsilon_0}$.
The centripetal force for circular motion is $F_c = \frac{mV^2}{r}$.

Setting $F_c = F_e$ yields $\frac{mV^2}{r} = K\frac{Ze^2}{r^2}$.
Multiplying both sides by $r$ gives $mV^2 = K\frac{Ze^2}{r}$.

Kinetic energy is $KE = \frac{1}{2}mV^2 = \frac{1}{2}K\frac{Ze^2}{r}$.

$KE = \frac{KZe^2}{2r}$ corresponds to option A.

Hence, the correct answer is option: 'A'.

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