Question

find the rms velocity of smoke particle of mass 1.23×10 to the power _27 kg at 300 kelvin? ​

Answer 1

Answer:

Correct option is

B

1.5×10

−2

m s

−1

According to Kinetic theory,

average K.E. of a gas molecule =

2

1

mv

rms

2

=

2

3

k

B

T

∴v

rms

=

m

3k

B

T

=

5×10

−17

3×1.38×10

−23

×273

=1.5×10

−2

ms

−1