The work done by the field of a nucleus in a complete circular orbit of the electron is |
$\frac{k (Ze) e}{r}$ $\frac{k (Ze) e}{r^2}$ $-\frac{k (Ze) e}{r^2}$ Zero |
Zero |
The correct answer is Option (4) - Zero The nucleus exerts an electrostatic force on the orbiting electron. This force acts as the centripetal force and always point towards the centre of circular orbit - $W=\vec F.\vec d=Fd\cos θ$ for a circular orbit, the force is always perpendicular to the instantaneous displacement of electron. $θ=90°$ |