# Statements regarding Brownian motion experiment

1. Nov 25, 2013

### coconut62

1. The problem statement, all variables and given/known data

In an experiment to demonstrate Brownian motion in a gas, a brightly illuminated cell containing smoke is viewed under a microscope. The observer sees a large number of bright specks undergoing random motion.

A) Light is being scattered by gas molecules.
B) The larger the smoke particle, the greater is the speed of the bright specks.
C) The lower the smoke pressure of the gas, the more frequent are the direction changes of the bright specks
D) The higher the temperature of the gas, the faster is the motion of the bright specks.

2. Relevant equations

None

3. The attempt at a solution

I chose D, the answer is D.

D is obviously correct.
I know B is incorrect.

But I am not very sure about A and C.

"A) Light is scattered by gas molecules" Isn't that so? Light is reflected in many directions by many gas molecules so it is scattered.

"C) The lower the smoke pressure, the more frequent are the direction changes of the bright specks" If smoke pressure is lower, it means there are fewer smoke particles. But how does that affect the rate of direction changes?

2. Nov 25, 2013

### Staff: Mentor

Why has the experimenter introduced smoke into the jar? What else is in the jar apart from smoke? (Smoke is not dirty gas molecules.)

3. Nov 26, 2013

### coconut62

The experimenter wanted to observe the motion of the smoke particles?
Apart from smoke there's air molecules in the jar.

Regarding my concern about choice A), I think the gas molecules means air molecules?

4. Nov 26, 2013

### ehild

Yes, it might be air or some other gas, but not the smoke. The smoke particles perform random "Brownian" motion as the gas molecules hit them in random directions. See, for example, http://en.wikipedia.org/wiki/Brownian_motion and the experiment dust particles in air here

The light might be scattered by the air molecules, too, but scattering is most effective when the wavelength of light is comparable to the size of the particle. That is true for the dust particles, so we see them bright as the light scattered from them reaches our eyes.
As for C) The number of the smoke particles in a given volume is much lower than the number of the gas molecules. The direction of motion of a bright speck (dust particle) changes if it is hit by a gas molecule. Of course, two dust particles can also collide but it is much less frequent than collision with air molecules.

ehild

Last edited by a moderator: Sep 25, 2014
5. Nov 26, 2013

### Staff: Mentor

It is not necessary that the gas be air here.

Although we know how light can be scattered by gas molecules, this experiment is not a demonstration of light being scattered by gas molecules. So that is why (A) is not the most/best correct answer.

6. Nov 26, 2013

### coconut62

Do you mean that when there are less smoke particles, the probability of one particle being bombarded by gas molecules increase so its change of speed increases?

7. Nov 26, 2013

### haruspex

No, ehild means that the gas molecules so vastly outnumber the smoke molecules that collisions between smoke molecules are irrelevant.
I would add that for the same reason, what the gas molecules do, in aggregate, is barely affected by the smoke molecules.
Therefore what you observe for one smoke molecule is unrelated to the number of smoke molecules.

8. Nov 26, 2013

### ehild

No, I did not mean so. Haruspex explained exactly what I meant
By the way, it is strange to speak about "smoke pressure" in the experiment.

ehild

Last edited: Nov 26, 2013