The velocity of the electron moving around the proton in a hydrogen atom, in an orbit of radius 5.3×10⁻¹¹ m, is 2.2×10⁶ m s⁻¹. Calculate the angular frequency of the electron moving around the proton.

Question

Accepted Answer

(4) 4.15×10¹⁶ rad s⁻¹. Given: v=2.2×10⁶ m s⁻¹ and r=5.3×10⁻¹¹ m. Idea: in uniform circular motion v=ω r, so ω=v/r. ω=(2.2×10⁶)/(5.3×10⁻¹¹)=0.415×10¹⁷ rad s⁻¹. ω=4.15×10¹⁶ rad s⁻¹. The ordinary frequency, in revolutions per second, is smaller by 2π: 
u=ω/(2π)=6.6×10¹⁵ Hz. That is not the angular frequency asked for.

The velocity of the electron moving around the proton in a hydrogen atom, in an orbit of radius 5.3×10⁻¹¹ m, is 2.2×10⁶ m s⁻¹. Calculate the angular frequency of the electron moving around the proton.

Step-by-step solution

Why the other options are wrong