The Bohr – deBroglie Connection

According to Bohr theory and deBroglie’s matter wave notion, how fast is an electron traveling around the nucleus when it is in the n=1 energy level? What is the electron’s kinetic energy?

The circumference, C, is 2pR

C = 2pR = 2px5.3x10^-11 m

Where R is from the Bohr radius for n=1

C = 3.33x10^-10 m

To be in phase, the wave must make an integral number of complete cycles around the path. For the first energy level, 1 complete cycle must be completed around the circumference of the n=1 path.

or

E= ½ mv² = ½ x 9.11x10^-31kg x (2.18x10^-6)² = 2.17x10^-18 J

This answer matches the energy of the n=1 very closely from the equation:

To test this we can see if the physical forces are balanced, that is:

centripetal force = force of attraction.

F_centripetal=mv²/R = F_attraction= kq²/R²

Where k is the electric field constant 9x10⁹ Nm²/C² and q is the charge on both an electron and a proton, 1.6x10^-19C.

F_centripetal=mv²/R =9.11x10^-31 kg x (2.18x10⁶ m/s)²/5.3x10^-11 m = 8.2x10^-8 Newtons

F_attraction= kq²/R² = 9x10⁹ Nm²/C²x(1.6x10^-19C)²/(5.3x10^-11)² m =8.2x10^-8 Newtons

Perfect! This should also work quite nicely for all hydrogen-like ions, He⁺¹, Li⁺², …