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 = p^{2}/2m

Then uncertainty in K.E is

∆K =∆p^{2}_{x/2m}

Put equation (i) in above equation

∆K= Ћ^{2} /2m(∆x)^{2} (2)

As potential energy is given by

∆V= -1/4∏ε_{0 }Ze^{2}/∆x (3)

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

∆E= ∆K+∆V

= Ћ^{2} /2∏(∆x)^{2} –Ze^{2}/4∏ε_{0}∆x

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

∆E= Ћ^{2} /2mr^{2} –Ze^{2}/4∏ε_{0}r (4)

The Uncertainty in total energy will be minimum if

d(∆E)/dr=0 and d^{2(}(∆E)/dr^{2} is positive

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

d(∆E)/dr=0= – Ћ^{ 2}/mr^{3}+Ze^{2}/4π ε_{0}r^{2} (5)

For minimum value of ∆E

d(∆E)/dr=0= – Ћ^{ 2}/mr^{2}+Ze^{2}/4π ε_{0}r^{2}

or Ze^{2}/4π ε_{0}r^{2}= Ћ^{ 2}/mr^{3}

Or r=4π ε_{0} Ћ^{ 2}/me^{2 }_{ }(6)

Further differentiating equation (5), we get

d^{2}(∆E)/dr^{2}=3 Ћ^{ 2}/mr^{4}-2Ze^{2}/4π ε_{0}r^{3}

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

d^{2}(∆E)/dr^{2}

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

Hence, the radius of first orbit is given by

r=4π ε_{0}Ћ^{ 2}/me^{2}=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.