Determine the maximum angular speed of the electron in a stationary orbit of a hydrogen atom.

Step-by-step solution

The angular speed of an electron in the n -th orbit of a hydrogen atom is given by _n = v_n r_n . The velocity of the electron in the n -th orbit is v_n = v₁ n , where v₁ 2.18 10^6 m/s . The radius of the n -th orbit is r_n = r₁ n^2 , where r₁ 0.529 10⁻¹⁰ m . Substituting these into the expression for angular speed gives _n = v₁ r₁ n^3 . For the angular speed to be maximum, the principal quantum number n must be minimum, which is n = 1 for the ground state. Substituting n = 1 , the maximum angular speed is _ max =