In one of my earlier articles, I have discussed the one the applications of the Heisenberg uncertainty principle that is non-existence of electron in the nucleus. Let us discuss today the one more application of the Heisenberg uncertainty principle that is the determination of the radius of the Bohr’s first orbit. Let us start:

If ∆x and ∆p_x are the uncertainties in the simultaneous measurements of position and momentum of the electron in the first orbit, then from uncertainty principle

∆x∆p_x = Ћ

Where Ћ = h/2∏

Or    ∆p_x = Ћ /∆x                                                                 (1)

As kinetic energy is given as

K = p²/2m

Then uncertainty in K.E is

∆K =∆p²_x/2m

Put equation (i) in above equation

∆K= Ћ² /2m(∆x)² (2)

As potential energy is given by

∆V= -1/4∏ε₀ Ze²/∆x                                                      (3)

The uncertainty in total energy is given by adding equations (2) and (3), that is

∆E= ∆K+∆V

= Ћ² /2∏(∆x)² –Ze²/4∏ε₀∆x

If ∆x = r= radius of Bohr’s orbit, then

∆E= Ћ² /2mr² –Ze²/4∏ε₀r                                                   (4)

The Uncertainty in total energy will be minimum if

d(∆E)/dr=0 and d²⁽(∆E)/dr² is positive

Differentiating equation (4) w.r.t. r, we get

d(∆E)/dr=0= – Ћ ²/mr³+Ze²/4π ε₀r² (5)

For minimum value of ∆E

d(∆E)/dr=0= – Ћ ²/mr²+Ze²/4π ε₀r²

or                  Ze²/4π ε₀r²= Ћ ²/mr³

Or                       r=4π ε₀ Ћ ²/me² (6)

Further differentiating equation (5), we get

d²(∆E)/dr²=3 Ћ ²/mr⁴-2Ze²/4π ε₀r³

By putting value of r from equation (6) in above equation, we get positive value of

d²(∆E)/dr²

Therefore equation (4) represents the condition of minimum in the first orbit.

Hence, the radius of first orbit is given by

r=4π ε₀Ћ ²/me²=0.53 angstrom                                  (For H atom Z=1)

Put value of r in equation (4), we get

E_min= -13.6 e V

This value is same as determined by using Bohr’s theory.

Therefore, with the help of Heisenberg’s uncertainty principle, one can determine the radius of the Bohr’s first orbit.