The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is ( h = Planck's constant, m is the mass of electron and r is the radius of the orbit.)

Step-by-step solution

According to Bohr's quantization condition, the angular momentum of an electron in the n -th orbit is given by: mvr = nh 2 For the first Bohr orbit, n = 1 : mvr = h 2 The velocity of the electron is: v = h 2 mr The number of revolutions per second (frequency f ) is given by the ratio of velocity to the circumference of the orbit: f = v 2 r Substituting the value of v into the frequency equation: f = h 2 mr 2 r = h 4 ^2 mr^2 Answer: h / 4 ^2 mr^2