# What is the temperature at which the motion of particles theoretically ceases?

##### 3 Answers

I think that you will have to tell us that one........

#### Explanation:

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The temperature at which the motion of particles theoretically ceases is

#### Explanation:

Absolute zero computes to

At this temperature, both the enthalpy (heat content) and entropy (state of randomness or disorder) approach zero. Effectively, the molecules of the gas are slowing down towards being motionless.

Absolute zero also describes a gas reaching a temperature from which no more heat can be removed. Experiments have shown that molecules continue to vibrate at absolute zero.

There are more thoughts on this topic here:

https://www.thoughtco.com/what-is-absolute-zero-604287

**ATOMIC MOTION AT 0 K**

In regards to atoms, yes, the motion of particles will stop at

When we consider non-interacting atoms in the classical limit, the **equipartition theorem** gives for the *average per-particle kinetic energy*:

#<< K_(avg,"trans") >> -= K_(avg,"trans")/N = 3/2 k_BT# , in#"J/particle"# where the

#3# comes from the three cartesian directions (#x,y,z# ),#N# is the number of particles,#k_B# is the Boltzmann constant, and#T# is the temperature in#"K"# .

**MOLECULAR MOTION AT 0 K**

Molecules, on the other hand, have chemical bonds, which obviously means they can stretch and/or bend. To a first-order approximation, neglecting rotation at **simple harmonic oscillator** system, with energy

#E_(upsilon) = hnu(upsilon + 1/2) = ℏ omega (upsilon + 1/2)# ,where

#upsilon# is the vibrational quantum number,#h# is Planck's constant, and#nu# is the fundamental vibrational frequency in#"s"^(-1)# .

#omega# is the angular frequency, or#2pinu# , and#ℏ = h//2pi# is the reduced Planck's constant.

At

#E_0 = hnu(0 + 1/2) = 1/2hnu#

and all molecules continue to vibrate in their ground state at