Question

Why is the Heisenberg uncertainty principle not applicable for a bigger molecule?

Answer 1

Hint:The uncertainty principle has a close relationship with the wave nature of particles, which become too small for a large mass to be seen or have any effect. When we increase the momentum the uncertainty of the position decreases For the maximum position of certainty, the momentum should be very low.

Complete step-by-step solution:
The magnitude of the uncertainty can be written as follows,
\[\Delta x.\Delta p{\text{ }} > = h/\left( {4\pi } \right)\]
Now the \[\Delta p\] can be written as,
\[\Delta p = m\Delta v\]( for the slow-moving particles concerning the speed of the light.)
It is very difficult to determine experimentally. Heisenberg formulated this equation to fit Schrodinger's equation regarding the superposition of a wave function. The particles cannot be superpositioned.
The superposition velocity has got the wave functions to that distribution location.
The uncertainty is too small to notice. It only notices microscopic particles.
Here, In the Heisenberg uncertainty principle, the plank constant is very small like that the uncertainties in the position and momentum of small objects
A phenomenon like the atomic process and displacement are critically applicable. This is the reason why the Heisenberg uncertainty principle is significant only for the smaller particles. The Heisenberg uncertainty principle very accurately measures the momentum of the particle but it not well accurately measures the position of the particle.

Note:Heisenberg's uncertainty principle requires small uncertainty in the position.
By using the uncertainty principle we can measure both position and momentum of the particle simultaneously.
The Heisenberg uncertainty principles are used in the gravitational wave interferometer.