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 ( K =1 / 4 ₀ )

Step-by-step 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 Ze^2 r^2 with K = 1 4 ₀ . The centripetal force for circular motion is F_c = mV^2 r . Setting F_c = F_e yields mV^2 r = K Ze^2 r^2 . Multiplying both sides by r gives mV^2 = K Ze^2 r . Kinetic energy is KE = 1 2 mV^2 = 1 2 K Ze^2 r . KE = KZe^2 2r corresponds to opti