EXPERIMENT 17. To Determine Avogadro s Number by Observations on Brownian Motion. Introduction

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1 EXPERIMENT 17 To Determine Avogadro s Number by Observations on Brownian Motion Introduction In 1827 Robert Brown, using a microscope, observed that very small pollen grains suspended in water appeared to dance about in a random manner. At first it was thought that the grains were living organisms and that this was the cause of the random motion. However it was found that different types of small particles, no matter what their nature or origin, exhibited such a motion. The magnitude of the motion depended on several factors such as the size of the particle, the viscosity of the fluid in which it was suspended and the temperature of the fluid. Many observers noted that these particles appeared to behave in the same way as the molecules of an ideal gas should behave according to the Kinetic Theory. The explanation of Brownian Motion was first given by Einstein and Smoluckowski in about They assumed that the particles were being continually bombarded from all sides by the molecules of the fluid in which they were suspended. If the particle is large the number of molecules striking it at any instant will be large and because of the randomness of molecular motion the net force on the particle would be zero. If, however, the particle is very small the number of molecules striking it at any instant will be small. In this case there will be a considerable probability that the momentum of the molecules striking it from one side will be greater than those striking it from the opposite side so that a net force will act on the particle causing it to move. The motion of the particle through the fluid will be opposed by the viscosity of the fluid. As a result of these forces the particles will be observed to move in a random manner. Vertical distribution of Brownian particles (Sedimentation equilibrium) Sedimentation equilibrium is established when a suspension is undisturbed for some time. The forces which determine the vertical distribution of the Brownian particles are (1) the gravitational force, mg, which acts on the particles and, if unopposed, would tend to deposit them on the bottom of the container where their gravitational potential energy would be a minimum, (2) forces due 17-1

2 to the collision of the molecules of the fluid with the particles and which, if unopposed, would distribute the particles uniformly throughout the fluid. To determine the vertical distribution consider first a vertical column of gas particles at temperature T in dynamic equilibrium and consider the forces acting on a horizontal slice of thickness dh (Fig. 17.1). Figure 17.1: Sedimentation equilibrium The force acting down is its weight A n m g dh where n is number of particles per unit volume at h and m is the mass of a particle. The force acting upwards is due to the difference in pressure acting on lower and upper faces of the slice. A dp Since the slice is in equilibrium the negative sign arises because dh dp A dp = A n m g dh (17.1) is obviously negative. Using the ideal gas law and Boltzmann constant k = R N 0 P V = RT = N 0 kt where N 0 is Avogadro s number p = N 0kT = nkt (17.2) V where n, as before, is the number of particles per unit volume. Therefore for an isothermal change dp = k T dn (17.3) 17-2

3 Therefore combining Eqs and 17.3 dn n = mg kt dh Integrating between heights h = 0 and h where the densities of particles are n 0 and n respectively. n dn n 0 n = mg kt h dh 0 ln n ln n 0 = mg kt h Therefore n = n 0 exp mg kt h (17.4) This expression gives the variation of the number of particles per unit volume as a function of height. This equation can be applied to Brownian particles of uniform size if the weight mg is replaced by the effective weight of a particle in a fluid taking buoyancy into account. Assuming the particles are spherical with radius a: Effective weight = 4 3 πa3 ρg 4 3 πa3 ρ g = 4 3 πa3 g(ρ ρ ) where ρ and ρ are the densities of the material of the particles and the fluid respectively. Therefore Eq becomes n = n 0 exp 4πa3 g(ρ ρ )N 0 h (17.5) 3RT where k has been replaced by R N 0. Jean Perrin carried out a classical series of experiments between 1900 and 1912 in which he verified Eq and determined a value for Avogadro s number. He was awarded the Nobel Prize in 1926 for this work. Experimental Procedure The particles used in the present experiment are polystyrene spheres of uniform size suspended in water. A drop of the suspension is placed on a microscope slide and is sealed in by means of a cover glass to prevent evaporation. When the suspension is first placed on the microscope slide the particles are uniformly distributed but after a number of days equilibrium is established and the density of the particles decreases as you come up through the suspension. The suspension is viewed with a high power microscope. Because of the small depth of focus only a thin layer ( 1 micron) of the suspension will be in focus at a time and the grains in that layer will be sharply defined. By raising or lowering the microscope by means of the focus controls one can bring layers at different depths into focus. There is a coarse and a fine focus control on the microscope: IN USING THE COARSE FO- CUS CONTROL CARE MUST BE TAKEN NOT TO DRIVE THE MICROSCOPE OBJECTIVE 17-3

4 THROUGH THE COVER GLASS. THIS IS BEST DONE BY ALWAYS OBSERVING THE OB- JECTIVE DIRECTLY WHENEVER THE COARSE FOCUS CONTROL IS USED TO RAISE THE MICROSCOPE STAGE. The fine focus control is marked off in divisions and this is used to measure the difference in height between different layers in the suspension. However the scale on the fine focus control must first be calibrated. This can be done by measuring the thickness of a glass slide in terms of the divisions on the fine focus control and then measuring the thickness of the slide by means of a micrometer screw. In using the fine focus control to measure depth care must be taken to avoid errors due to backlash which arises from looseness in the mechanical parts of the microscope. To avoid this error it is best to measure depth while raising the microscope stage. When determining the density of the particles, the light source should be adjusted to give the smallest depth of focus and should not be altered for the rest of the experiment. Because of the random motion of the particles it is not possible to determine the density of the particle at a given level in a straight forward way i.e. by counting all the particles in the field of view. One way of determining the density is to make use of the small squares in the eye-piece. Levels should be chosen so that the average number of particles appearing in the squares is less than five. In that case the number of particles in a square can be counted at a glance. For each level a suitable number of estimates of the number of particles per square should be made, the mean of these estimates will then give a reliable value for the density at the level in question. To make the estimates independent the observer should make successive estimates in different squares and in an unbiased way. To investigate what constitutes a suitable number of estimates it should be noted that the estimates at a given level are distributed according to the Poisson Distribution (see Appendix). Because of this the standard error on the value of the density n at any level depends not on the number of estimates made but on the total number of particles counted. The total number of particles counted at a given level, ne, (where n = mean number in a square and E = number of estimates) also obeys the Poisson distribution. Therefore, because the standard deviation of a Poisson distribution is equal to the root of the mean, the standard error on ne is ne ± ne and the fractional error ne ne = 1 ne It follows that if for example you make 200 estimates at a level where n 1, then at a level where n 5, 40 estimates will be sufficient to give the same accuracy. Taking the natural log of Eq ln n = 4πa3 g(ρ ρ )N 0 3RT h + ln n 0 it is seen that if ln n is plotted against h a straight line should result with, slope = 4πa3 g(ρ ρ )N 0 3RT Since the density of polystyrene, ρ, is kg/m 3 Avogadro s number can be determined from this expression given the diameter of the particle to be 0.944µ m. 17-4

5 Avogadro s number is a fundamental physical constant giving the number of molecules in a mode of a substance. Combining it with values of atomic weight, atomic masses can be determined and combining it with Faraday s constant a value can be obtained for the fundamental unit of charge. NOTE : WHEN USING OPTICAL INSTRUMENTS THE GLASS SURFACES MUST NEVER BE TOUCHED WITH THE FINGERS SINCE THIS WILL LEAVE A FINGER PRINT WHICH WILL IMPAIR THE QUALITY OF THE IMAGE. LENS TISSUE SHOULD ALWAYS BE USED TO CLEAN THE GLASS SURFACES. Data Analysis : Print out each image taken and attach to report. Questions 1. How important was Perrin s experiment on Brownian motion in the development of Physics? 2. Under what conditions would a count of all the particles in the field of view give a reliable estimate of the density? How could these conditions be achieved? 3. Suppose you make 200 estimates of the number of particles in the small circle and find the mean number to be 5. What is the error on this mean? 4. How will the Brownian motion of a particle vary with (a) the size of the particle, (b) the viscosity of the fluid and (c) the temperature? Reference Physics of Atom, Wehr-Richards, Chapter 1. Appendix The Poisson Distribution As was said in the text the individual estimates of the number of particles per square at a given level are distributed according to the Poisson distribution. To prove this we will calculate the probability of observing x particles in a square at a given level if the mean number in a square at that level is N. If we take a square and divide it up into a large number, m, of sub-squares of equal area, Fig. 17.2, then because the Brownian particles are moving at random, the probability of finding a particle in a sub-square is p = N m (17.6) and is the same for all sub-squares. The probability of two Brownian particles in a sub-square is ( nm ) 2 since the two events are independent and is negligible if m is large. The probability of not getting a particle in a sub-square is (1 p) (17.7) 17-5

6 Figure 17.2: A grid similar to that used in the experiment One way of getting x particles in a square is to get a particle in each of the first x sub-squares and none in the remaining (m x) sub-squares. The probability for this would be p x (1 p) m x (17.8) Another way would be to get no particle in the first sub-square, a particle in each of the next x sub-squares and none in the remaining (m-x-1) sub-squares. The probability for this arrangement would also be given by Eq The above are just two of m Cx mutually exclusive ways of obtaining x particles in a square. Therefore the probability of obtaining x particles in a square is, using the summation law for mutually exclusive events, P (x) = m Cxp x (1 p) m x = m! x!(m x)! px (1 p) m x This is the Binomial distribution. We want to find the limit of this distribution as m and therefore p 0 since (Eq. 17.6) pm = N, the mean number of counts per square which is a constant for a given level. Now we can write m(m 1)(m 2)... (m x + 1) P (x) = p x (1 p) m x x! Multiplying above and below by m x we get P (x) = = m(m 1)(m 2)... (m x + 1) x!m x (mp) x (1 p) m x ( 1 1 ) ( 1 2 ) (... 1 x + 1 ) m m m x! (1 p)m x since mp = N = ( ) ( ) ( ) 1 1 m 1 2 m... 1 x + 1 m ( (1 p) x (1 p) 1/p ) N x! 17-6

7 If we take the limit of this expression as m which also implies p 0 since mp = N a constant, we get p(x) = Nx x! (1 p) Lim 1/p N p 0 since = Nx x! e N Poisson Distribution Lim p 0(1 + p) 1/p = e The mean and standard deviation of the Poisson Distribution Using the expansion e x = 1 + x + x2 2! + x3 3! + x4 4! +... it is easy to show that Poisson distribution is normalised and that its mean, as you would expect, is N. By definition the standard deviation is σ 2 = = = N 2 x! e N (N x) 2 x! e N (N 2 2Nx + x 2 ) x! e N 2N = N 2 2N 2 + x! e 2 x! e + x! e 2 = N 2 + e N (N + 22 N 2 2! + 32 N 3 3! + 42 N 4 4! +...) = N 2 + e N (N + 2N 2 + 3N 3 The terms can be rearranged in the following way σ 2 = N 2 + e N ( N(1 + N + N 2 = N 2 + e N (Ne N + N 2 e N ) 2! + N 3 3! 2! N 4 3! +...) +...) + N 2 (1 + N + N 2 2! + N 3 3! +...) )

8 since the series in brackets are the expansion of e N. σ 2 = N 2 + N + N 2 = N therefore σ = N Therefore the standard deviation of a Poisson Distribution is equal to the square root of the mean. P(x) is plotted against x for a number of values of N in Fig It is seen that for large N the distribution is almost symmetrical about the mean but as N decreases the distribution becomes more and more skew. The dotted curve is a normal distribution with mean 10 and standard deviation 10 and it is a fairly good fit to the Poisson distribution with the same mean. This fit improves as N increases so that for values of N greater than 10 the Poisson distribution can be approximated by the Normal Distribution with the same mean and a standard deviation equal to root of the mean. Figure 17.3: P(x) as a function of x for a number of values of N 17-8