According to **Heisenberg’s uncertainty principle**, macroscopically it is possible to exactly measure the position of moving particles at any instant and the momentum of the particles at that position but microscopically it is not possible to measure exactly the position of a particle and its momentum simultaneously. The reason is that the behavior of moving a particle is expressed by a wave packet anywhere with the group velocity.

According to Max-Born, particles can be found anywhere within the wave packet i.e. position of a particle is uncertain within the limits of the wave packet. For an extended wave packet, the momentum of the particle can be exactly measured but its position cannot be determined exactly. On the other hand, if the wave packet is small, the momentum of the particle cannot be determined with certainty, but its position can be determined.

## Statment:

**”It is impossible to measure or calculate exactly, both the position and momentum of a particle”**

According to Heisenberg’s uncertainty principle, the product of uncertainties in the measurement of momentum and position particle is of the order of h/4π. If the uncertainty in the measurement of position is ∆x, which of momentum is ∆p, then the Heisenberg uncertainty principle is:

**∆x ∆p ≥ h/4π**

Where,

- ∆x = Uncertainty in position
- ∆p = Uncertainty in momentum
- h = Planck’s constant

The uncertainty principle is applicable only for microscopic particles like electrons, protons, etc., and for macroscopic particles it has no meaning. Compton’s effect can help us understand the uncertainty principle. If we determine the position of microscopic particles like the electron, visible light cannot help us. It is because the wavelength of visible light is billionth of a trillionth time large as compared to the diameter of the electron. So, we have to use the light of a shorter wavelength (X-rays). When this photon of X-rays strikes an electron, the momentum of the electron will change. In other words, due to the change of velocity of the electron, the uncertainty of momentum will appear. As

Photon Energy ∝ hc/λ

Photon Energy ∝ 1/λ

The smaller the wavelength of X-rays, the greater will be the energy of the photon. Hence, the collision of X-rays with electrons will bring about greater uncertainty in momentum. So, an attempt to determine the exact position of an electron has rendered its momentum uncertain. When we use the photon of a longer wavelength to avoid the change in momentum, the determination of the position of the electron becomes impossible.

## Derivation of Heisenberg’s Uncertainty Principle:

### Group Velocity:

The group velocity of a wave is the velocity with which the variation in the shape of the wave’s amplitude (known as the modulation or envelope of the wave) propagates through space.

From group velocity for wave packet,

V_{g} = ∆ω/∆K = ω_{1}− ω_{2} / K_{1}− K_{2} = dω/dK ——– (1)

Where,

- ω = Angular velocity
- K = Propagation constant
- Value of propagation constant K = 2π/λ

K = 2π/λ

dK = d(2π/λ)

dK = 2π d( λ )^{−1}

dK = −2π λ^{−2}dλ

dK = −2π dλ / λ^{2}

Put dK in eq (1)

V_{g} = dω λ^{2} / −2π dλ ——– (2)

From de-Broglie wave particle duality,

λ = h/p ——– (3)

dλ = d(h/p)

dλ = h d(p)^{−1 }

dλ = -h dp / p^{2} ——– (4)

Put eq (3) and (4) in eq (2)

V_{g} = −dω/2π × (h^{2}/p^{2}) × 1/(−h dp/p^{2} )

V_{g} = dω/2π × (h^{2}/p^{2}) × p^{2}/(h dp )

V_{g} = h/2π × dω/dp ——– (5)

Velocity can be expressed as, V = dx/dt (change in position w.r.t time)

V = dx/dt = ∆x/∆t ——– (6)

Compare eq (5) and (6)

∆x/∆t = h/2π × ∆ω/∆p

∆x ∆p = h/2π × ∆ω ∆t——– (7)

For least time period the measurement of angular frequency is,

∆ω ≥ 1/∆t

From eq (7)

∆x ∆p ≥ h/2π × 1/∆t × ∆t

**∆x ∆p ≥ h/2π OR ∆x ∆p ≥ h/4π**

## Some other forms of Heisenberg’s Uncertainty Principle:

∆p ∆x ≥ h/4π ——– (A)

As momentum is the product of mass and velocity,

p = mv

∆p = m∆v

put the value of ∆p in eq (1)

m ∆v × ∆x ≥ h/4π

**∆v ∆x ≥ h/4πm**

Divided by ∆t from eq (A)

∆p/∆t × ∆x ∆t ≥ h/4π

- ∆p/∆t = Change in momentum w.r.t time = Force
- F × ∆x = Force into displacement = Energy

**∆E ∆t ≥ h/4π**

### Prove that ∆λ ∆x ≥ λ^{2} / 4π?

From de-Broglie,

λ = h/p

p = h/λ

p = hλ^{-1}

dp/dλ = h d λ^{-1}/dλ

dp/dλ = − h λ^{-2}

∴ we do not consider a negative sign

dp/dλ = h λ^{-2}

∆p/∆λ = h λ^{-2}

∆p = h λ^{-2} ∆λ

put the value of ∆p in eq (A)

h λ^{-2} ∆λ ∆x ≥ h/4π

λ^{-2} ∆λ ∆x ≥ 1/4π

∆λ ∆x ≥ 1/4π λ^{-2}

**∆λ ∆x ≥ λ ^{2}/4π**

### Prove that ∆K.E ∆x ≥ hv/4π?

K.E = 1/2×mv^{2}

d (K.E) /dv = 1/2 m d (v^{2}) /dv

d (K.E) /dv = 1/2 m × 2 V^{2-1}

d (K.E) /dv = mv

∆ (K.E) /∆v = mv

∆ (K.E) = m ∆v × v

∴ ∆p = m ∆v

∆ (K.E) = ∆p v

∆p = ∆ (K.E) / v

put ∆p in eq (A)

∆ (K.E) / v × ∆x ≥ h/4π

**∆ (K.E) × ∆x ≥ hv/4π**

## Is electron reside inside the nucleus? If not why?

If we assume that the electron is inside the nucleus, then since the radius of the nucleus is nearly 10-14m, therefore the maximum uncertainty in the position of an electron to be inside the nucleus is,

(∆x)_{max} ≈ 2r ≈ 2×10^{-14}

From Heisenberg uncertainty principle,

∆p ∆x ≥ h/4π

(∆p)_{min} ≈ h/4π(∆x)_{max}

(∆p)_{min} ≈ 6.63×10^{-34}/4 × 3.14 × (2×10^{-14})

(∆p)_{min} ≈ 5.275×10^{-21} Kg m/s

Thus, the momentum of the electron must be more than 5.275×10^{-21} Kg m/s. The relativistic energy of the electron is,

E = √p^{2} c^{2}× m_{°}^{2} c^{4} ≈ pc

Assuming the momentum of the electron is 5.275×10^{-21} Kg m/s, Its energy

E ≈ 5.275×10^{-21} × 3.00×10^{8}

E ≈ 9.89 MeV

Thus, if we assume that the electron resides inside the nucleus, its energy must be nearly 9.89 MeV, where the maximum energy of β particles or electrons emitted from the nucleus is nearly 2 to 3 MeV. Thus, we conclude that electrons cannot reside inside the nucleus.

## Problems related to uncertainty principle:

### Problem 1:

If uncertainty in the position of an electron is 10^{-8} cm. Calculate the uncertainty in its velocity?

∆x = 10^{-8} cm = 10^{-8}/100 m = 10^{-10}

∆p ∆x ≥ h/4π

m∆v × ∆x ≥ h/4π

m∆v × ∆x = h/4π (limiting value)

- mass in Kg
- ∆v in m/s
- ∆x inn m

9.1×10^{-31} Kg × ∆v × 10^{-10} m = 6.63×10^{-34} / 4×3.14

∆v = 6.63×10^{-34} / 4×3.14 × 9.1×10^{-31} × 10^{-10}

**∆v = 5.65×10 ^{5} m/s**

### Problem 2:

If uncertainty in the position of an electron is 1/3 pm. what is uncertainty in velocity?

∆x = 1/3 pm = 1/3 × 10^{-12} m

∆p ∆x ≥ h/4π

m∆v × ∆x ≥ h/4π

m∆v × ∆x = h/4π (limiting value)

9.1×10^{-31} Kg × ∆v × 1/3 × 10^{-12} m = 6.63×10^{-34} / 4×3.14

∆v = 6.63×10^{-34} / 4×3.14 × 9.1×10^{-31} × 1/3 × 10^{-12}

**∆v = 1.6×10 ^{8} m/s**

### Problem 3:

If uncertainty in the measurement of position and momentum of an electron is equal. Calculate the uncertainty in the measurement of velocity?

Given, ∆x = ∆p

∆p ∆x ≥ h/4π

∆p ∆p = h/4π

(∆p)^{2} = h/4π

∆p = √h/4π

m ∆v= √h/4π

**∆v= 1/m × √h/4π**

### Problem 4:

If uncertainty in the measurement of position and velocity of an electron is equal. Calculate the uncertainty in the measurement of momentum?

Given, ∆x = ∆v

∆p ∆x ≥ h/4π

m ∆v × ∆v = h/4π

(∆v)^{2} = h/4πm

∆v = √h/4πm

Multiply by ”m” both side

m ∆v = m √h/4πm

∆p= m √h/4πm

**∆v= √hm/4π**

### Problem 5:

If the electron is to be located within 5×10^{-5} Angstrom. What will be the uncertainty in the velocity? Can an electron remain inside a nucleus?

∴ 1 Angstrom = 10^{-10} m

5×10^{-5} A = 5×10^{-5} × 10^{-10} = 5×10^{-15} m

∆p ∆x ≥ h/4π

m∆v × ∆x ≥ h/4π

m∆v × ∆x = h/4π (limiting value)

9.1×10^{-31} Kg × ∆v × 5 × 10^{-15} m = 6.63×10^{-34} / 4×3.14

∆v = 6.63×10^{-34} / 4×3.14 × 9.1×10^{-31} × 5×10^{-12}

**∆v = 1.2×10 ^{-10} m/s**

Which is greater than speed of light, ∆v > c

The uncertainty in velocity is more or greater than the speed of light. Hence, the electron cannot remain inside the nucleus.

## Multiple Choice Question’s related to Uncertainty Principle:

Q1: If uncertainty in the position of an electron is zero, the uncertainty in its momentum will be:

a) < h/4π

b) > h/4π

c) 0

d) ∞

**D ∆p ∆x ≥ h/4π If ∆x—->0 then ∆p—->∞**

Q2: If uncertainty in position and momentum are equal then the uncertainty in velocity is:

a) √h/2π

b) √h/π

c) 1/2m √h/π

d) 1/m √h/π

**C ∆p ∆x ≥ h/4π** —-> **∆x = ∆p so we can write (∆p) ^{2} = h/4π —-> ∆p = √h/4π —-> m∆v = √h/4π —-> ∆v = 1/2m √h/π **

Q3: Calculate the uncertainty in the position of a particle when uncertainty in momentum is 1×10^{-3} g cm/sec:

a) 0.527×10^{-24} cm

b) 0.302×10^{-24} cm

c) 1.325×10^{-23} cm

d) 1.206×10^{-23} cm

**A ∆p ∆x ≥ h/4π so, ∆x = h/4π∆p —-> ∆x = 6.63×10 ^{-27} / 4×3.14×10^{-3} —-> ∆x = 0.527×10^{-24} cm**

Q4: What will be the uncertainty in the velocity of a cricket ball if its uncertainty in position is of the order of 1A^{°} (if m = 0.15 Kg)

a) 2.24×10^{-24}

b) 3.51×10^{-24}

c) 2.24×10^{-23}

d) 3.51×10^{-23}

**B ∆p ∆x ≥ h/4π —-> m∆v = h/4π∆x —-> ∆v = h/4πm∆x —-> ∆v = 6.63×10 ^{-34} / 4×3.14×0.15×10^{-10} —-> 3.51×10^{-24} m/s**

Q5: Which of the following contradicts with Heisenberg uncertainty principle?

a) Bohr’s Theory

b) Mole

c) de-Broglie

d) All

**A Bohr’s Theory**