## Applications of the Heisenberg Uncertainty Principle: The Radius of Bohr’s First Orbit

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

∆x∆px = Ћ

Where Ћ = h/2∏

Or    ∆px = Ћ /∆x                                                                 (1)

As kinetic energy is given as

K = p2/2m

Then uncertainty in K.E is

∆K =∆p2x/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 Continue reading “Applications of the Heisenberg Uncertainty Principle: The Radius of Bohr’s First Orbit”

## Heisenberg uncertainty principle

Statement: According to Heisenberg uncertainty principle, it is impossible to measure the exact position and momentum of a particle simultaneously within the wave packet.

We know, group velocity of the wave packet is given by

vg =∆ω/∆k

Where ω is the angular frequency and k is the propagation constant or wave number

But vg is equal to the particle velocity v

Thus vg = v =  ∆ω/∆k                                                   (1)

But                    ω=2пf

Where f is the frequency

Therefore   ∆ ω = 2п ∆ f                          (2)

Also                       k=2 п/λ

Since         de-Broglie wavelength   λ=h/p

By putting this value in equation of k, we get

k=2пp/ λ

Therefore                ∆k=2п∆p / λ                                 (3)

Put equations (2) and (3) in equation (1), we get

v= 2пh∆f/2п∆p =h∆f /            (4)

Let the particle covers distance ∆x in time ∆t, then particle velocity is given by

v  = ∆x/∆t                     (5)

Compare equations (4) and (5), we get

∆x/∆t=h∆f/∆p

Or                          ∆x.∆p=h∆f ∆t                                     (6)

The frequency ∆f is related to ∆t by relation

∆t≥ 1/∆f                                           (7)

Hence equations (6) becomes

∆x.∆p≥ h

A more sophisticated derivation of Heisenberg’s uncertainty principle  gives

∆x.∆p=h/2п                                          (8)

Which is the expression of the Heisenberg uncertainty principle.

As the particle is moving along x-axis. Therefore, the momentum in equation (8) of Heisenberg’s uncertainty principle should be the component of the momentum in the x-direction, thus equation Heisenberg’s uncertainty principle can be written as,

∆x.∆px=h/2п                             (9)

Note: There can not be any uncertainty if momentum is along y direction.

Q: Why there is uncertainty in position and momentum?

Answer: Because the particle is always in disturbed state during motion. It is not possible to calculate the position and momentum of particle simultaneously.