Question

Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of the hydrogen atom.

Answer 1

The radius of the nth Bohr orbit is

r = `(ε_0"h"^2"n"^2)/(pi"mZe"^2)`    ....(1)

and the linear speed of an electron in this orbit is

v = `"Ze"^2/(2ε_0"nh")`     ....(2)

where ε₀ ≡ permittivity of free space, h =Planck's constant, n ≡ principal quantum number, m ≡ electron mass, e ≡ electronic charge and Z = the atomic number of the atom.

Since angular speed ω = `"v"/"r"`, then from Eqs. (1) and (2), we get, 

ω = `"v"/"r" = "Ze"^2/(2ε_0"nh") * (pi"mZe"^2)/(ε_0"h"^2"n"^2) = (pi"mZ"^2"e"^4)/(2ε_0^2"h"^3"n"^3)`    ...(3)

which gives the required expression for the angular speed of an electron in the nth Bohr orbit. From Eq. (3), the frequency of revolution of the electron,

f = `omega/(2pi) = 1/(2pi) xx (pi"mZ"^2"e"^4)/(2ε_0^2"h"^3"n"^3) = ("mZ"^2"e"^4)/(4ε_0^2"h"^3"n"^3)`     .....(4)

Obtain the formula for w and continue as follows:

ω(maximum) = `(pi"me"^4)/(2ε_0^2"h"^3)` (for Z = 1 and n = 1)

`= ((3.142)(9.1 xx 10^-31 "kg")(1.6 xx 10^-19"C")^4)/((2)(8.85 xx 10^-12"C"^2//"N.m"^2)^2(6.63 xx 10^-34 "J.s")^3)` rad/s

`= ((3.142)(9.1)(1.6)^4(10^-107))/((2)(8.85)^2(6.63)^3(10^-126))`

= 4.105 × 10¹⁶ rad/s