Numerical problems on Heisenberg’s uncertainty principle 

According to Heisenberg’s uncertainty principle, there is uncertainty while measuring a particle’s variable. The principle, which is commonly applied to a particle’s position and momentum, states that the more specifically a position is known, the more uncertain the momentum is, and the same happens when the momentum is known precisely and the position is uncertain. 

This is in contrast to classical Newtonian physics, which states that with good enough equipment, all particle variables are traceable to an unlimited uncertainty. The Heisenberg uncertainty principle is a basic principle of quantum physics that explains why a scientist cannot simultaneously measure many quantum variables. Let us learn more about this with numerical questions on Heisenberg’s uncertainty principle and their answers in detail.

Heisenberg uncertainty principle formula

Werner Heisenberg, a German scientist, proposed this idea in 1927. This principle states that each particle’s position and momentum cannot be measured with infinitely high precision at the same time. The product of position and velocity uncertainty is equal to or larger than a very small physical quantity, h. As a result, this product of uncertainty will only be meaningful for atoms and subatomic particles with extremely small masses.

At all times, the value of position and momentum is higher than h/4π.

Formula:  ∆x∆p  ≥  h/4π

Where:

The Planck constant (6.62607004 x 10-34 kg m2 / s) is represented by h.

The uncertainty in momentum is denoted by the letter ∆p.

The uncertainty in position is denoted by the letter ∆x.

Heisenberg’s uncertainty principle formula can also be expressed as:

 ∆x∆mv ≥   h/4π

This is because momentum is p = mv.

When position or momentum are accurately measured, it immediately suggests a higher inaccuracy in the measurement of the other quantity.

Questions and solved answers for Heisenberg’s uncertainty principle

Example 1:  Let us suppose, for the sake of argument, that electrons exist in the nucleus. The diameter of the nucleus is about 10-14 metres. If the electron is to exist inside the nucleus, the position of the electron must be uncertain.

∆x= 10-14 m

According to the principle of uncertainty,

 ∆x∆px = h/2π

Therefore, ∆px = h/2π∆x

∆px =6.62 x10-34/2 x 3.14 x 10-14

∆px=1.05 x 10-20 kg m/ sec

If the uncertainty in the electron’s momentum is this, then the electron’s momentum should be at least of this order, p = 1.05×10-20 kg m/sec. With such a large momentum, an electron’s velocity must be similar to that of light. As a result, the following relativistic formula should be used to compute its energy:

E=  √( m20 c4 + p2c2)

E =  √[(9.1×10-31)2 (3×108)4 + (1.05×10-20)2(3×108)2]

= √[(6707.61×10-30) +(9.92×10-24)]

= √[(0.006707×10-24) +(9.92×10-24)]

= √(9.9267×10-24)

E = 3.15×10-12 J

Or , E = 19.6 MeV

As a result, if the electron occurs in the nucleus, its energy should be in the range of 19.6 MeV. However, beta particles (electrons) released from the nucleus during b-decay have an energy of around 3 MeV, which differs significantly from the value obtained of 19.6 MeV. The second reason an electron cannot exist inside the nucleus is that no electron or particle in the atom has an energy greater than 4 MeV, according to experimental evidence.

As a result, it has been established that electrons do not exist within the nucleus.

Example 2: Calculate the uncertainty in the momentum of an electron if uncertainty in its position is 1 Å ( 10-10 m).

According to Heisenberg’s uncertainty principle:

∆x∆p  ≥   h/4π

 ∆p ≥    h/4π∆x

∆p  ≥ 5.28 x 10–²5kg m s-1

Hence, the uncertainty in the momentum of the electron is 5.28 x 10–²5kg m s-1

 Conclusion

The Heisenberg uncertainty principle says that determining a particle’s position and velocity at the same time is impossible. The interaction of an electron with photons of light, for example, would be used to detect it. Because photons and electrons have roughly the same energy, using a photon to locate an electron will knock the electron off track, leaving the location of the electron unknown. 

Because of their mass, we don’t have to bother about the uncertainty principle with massive ordinary objects. If you use a flashlight to look for something, the photons from the flashlight will not cause the object you’re looking for to move.