Question

The value closest to the thermal velocity of a helium atom at room temperature (300 k) in ms1 is : [ kb=1.41023 j/k; m he=71027 kg ]

Answer 1

answer is 1.3×10∧3

detail is given

T=300k

Mhe=7×10∧-27kg

v=?

formula  Vrms=√3Rt/M

                      =√3×1.4×10^23×300/7×10^-27

            V          =√1.8×10^6≈1.3×10^3 m/s

        hope it is helpfull

                  =

Answer 2

The value closest to the thermal velocity of a helium atom at room temperature (300 k) in ms1 is : [ kb=1.41023 j/k; m he=71027 kg ]

Answer:

The solution is:

The values of the closet to the thermal velocity of a helium atom at room temperature are 300 k.

Vp = √(█(3KT@-@m))  

Vp = √(3*1.38*10)-23 *300

       ----------------------------

          7*10-27

we get = 1.3* 103