A hydrogen atom comprises one proton and one electron. We can assume that the electron revolves in a circle of radius 0.53 angstrom ( 1 angstrom = 10⁻¹⁰ m , abbreviated as Å) with the proton situated at the centre. Under these conditions, the hydrogen atom is considered to be in the ground state. Determine the speed of the electron in the ground state of a hydrogen atom.

Step-by-step solution

The centripetal force for the circular motion of the electron is provided by the electrostatic force of attraction between the proton and the electron. mv^2 r = 1 4 ₀ e^2 r^2 v = e^2 4 ₀ mr Substituting the standard values e = 1.6 10⁻¹⁹ C , m = 9.1 10⁻³¹ kg , r = 0.53 10⁻¹⁰ m , and 1 4 ₀ = 9 10^9 N m ^2/ C ^2 : v = 9 10^9 (1.6 10⁻¹⁹)^2 9.1 10⁻³¹ 0.53 10⁻¹⁰ v = 23.04 10⁻²⁹ 4.823 10⁻⁴¹ v = 4.77 10¹² v = 2.18 10^6 m/s Alternatively, using Bohr's quantization condition for the ground state ( n=1 ): mvr = h 2 v = 6.63 1