Tutorial questions and suggested solutions

1. Number of bacteria living with body of the average healthy adult human is estimated to outnumber human cells 10:1. Estimate mass of all the bacteria in your body (or body of your favourite pop-star). Use internet for additional information. Explain your quantitate reasoning. You can use internet for additional information that you might require (but not to get the exact number).

Assumptions: Average adult human weighs in the order of 102 kg. Average diameter of a human cell is about 10-5 m. Average diameter of bacteria is 10-6 m. Cell density of human cell and bacteria is similar to that of water 1kg/m3

Volume of one cell = 10-15 m3/cellMass of one cell = 10-15 kg/cellNumber of cells = 102 kg 10-15 kg/cell = 1013 cellSince number of bacteria outnumber human cells 10:1 1014 bacteriaVolume of one bacteria = 10-18 m3/bacteriaMass of one bacteria = 10-18 kg/bacteriaTotal mass of bacteria = 10-18 kg/bacteria 1014 bacteria = 10 -4 kg = 0.1g

2. Calculate the energy yield per liter of oxygen consumed for each food type, and note that it is roughly constant. Thus we can determine a person’s metabolic rate simply by measuring her rate of oxygen consumption. In contrast, the CO2/O2 ratios are different for the different food groups; this circumstance allows us to estimate what is actually being used as the energy source, by comparing oxygen intake to carbon dioxide output.

Food kcal/g liters O2/g liters CO2/g kcal/liters O2 CO2/O2

Carbohydrate 4.1 0.81 0.81 5.06 1Fat 9.3 1.96 1.39 4.74 0.71

Protein 4.0 0.94 0.75 4.26 0.80Alcohol 7.1 1.46 0.97 4.86 0.66

Based on the data given, the average energy yield per liter of O2 is 4.73 0.34 kcal. It is noted that for the CO2/O2 ratios, the average value is 0.79 0.15 which seem to be

more constant as compared to the average energy yield per liter of O2.

b. An average adult at rest uses about 16 liters of O2 per hour. The corresponding heat release is called the “basal metabolic rate” (BMR). Find it, in kcal/hour and in kcal/day.

16 liters/hr 4.73kcal/liter O2 = 75.7kcal/hr75.7kcal/hr 24 hr/day = 1820kcal/day

c. What power output does this correspond to in watts?

75.7kcal/hr 4184J/kcal 1hr/3600s = 88W

d. Typically, the CO2 output rate might be 13.4 liters/hour. What, if anything, can you say about the type of food materials being consumed?

13.4/16 = 0.8375. Although this value lies somewhere between carbohydrates and proteins, we cannot just simply assume that it is a combination of both. In actual fact for a typical meal, we often consume a mixture of carbohydrates, fat and protein (alcohol sometimes).

e. During exercise, the metabolic rate increases. Someone performing hard labor for 10 hours a day might need about 3500 kcal of food per day. Suppose the person does mechanical work at a steady rate of 50W over 10 hours. We can define the body’s efficiency as the ratio of mechanical work done to excess energy intake (beyond the BMR calculated in (b)). Find this efficiency.

The BMR calculated was found to be about 2000kcal/day. Thus the excess energy intake would be about 1500kcal. Body efficiency:(50J/s 10hr 3600s/hr) (1500kcal 4184J/kcal) = 0.287 0.3

3. Binomial distribution is defined as:

P (k )= N !k ! (N−k ) !

pk (1− p )N−k

Prove that it is normalised, i.e.

∑k=0

N

P (k )=1

Using the binomial theorem:

(a+b)N=∑k=0

N

(Nk )a(N−k)bk=∑k=0

N N !k ! (N−k )!

a(N−k)bk

Thus:

∑k=0

N

P (k )=¿∑k=0

N N !k ! (N−k ) !

pk (1−p )N−k=∑k=0

N

(Nk ) pk

(1−p )N−k ¿

Lettingq=1−p,

∑k=0

N

(Nk ) pk

qN−k=(p+q)N=1

The binomial distribution is normalized.

4. The continuous Gaussian distribution is given by:

ρ ( x )= 1σ √2π

e−( x−μ )2

2σ2

wherex∈[−∞ ,∞ ]. Gaussian integral formula is given by:

I (b )=∫−∞

∞

e−b y2dx=√ πb

a. Please show that:

ddb

I (b )=∫−∞

∞

y2 e−b y2dx

Using Leibniz integral rule:

ddx (∫a

b

f ( x , t )dt )=∫a

b ∂∂x

f ( x , t )dt

ddb

I (b )=∫−∞

∞ ∂∂b

e−b y2dy=∫−∞

∞

− y2e−b y2dy=−∫−∞

∞

y2 e−b y2dx

b. Prove that:

∫−∞

∞

ρ ( x )dx=1∧ ⟨ x ⟩=μ

Let:

z=

(x−μ )σ √2∧dz

dx= 1

σ √2Then:

∫−∞

∞

ρ ( x )dx=∫−∞

∞ 1σ √2π

e−( x−μ)2

2σ 2 dx=∫−∞

∞ 1√ π

e− z2dz= 1√π ∫−∞

∞

e− z2dz

Using:

(∫−∞

∞

e− x2dx)2

=∫−∞

∞

e−x2dx ∙∫−∞

∞

e− y2dy=∫−∞

∞

∫−∞

∞

e−(x2+ y2)dxdy

Converting into polar coordinates where x2+ y2=r2 anddxdy=rdrdθ:

∫−∞

∞

∫−∞

∞

e−( x2+ y2 )dxdy=¿∫

0

2π

∫0

∞

e−(r2 ) rdrdθ=2 π ∙∫

0

∞

r ∙ e−r2dr ¿

Let:

u=−r 2∧dudr=−2 r

2π ∙∫0

∞

r ∙e−(r2 )dr=2π ∙∫

−∞

0 12

eu du=π ∙ (e0−e−∞ )=π

Thus:

(∫−∞

∞

e− z2dz )2

=π∧∫−∞

∞

e− z2dx=√π

∴∫−∞

∞

ρ ( x )dx= 1√ π∫−∞

∞

e−z2dz= 1√π

∙√π=1

The mean of a continuous distribution function is given by:

⟨ f ( x ) ⟩=∫a

b

f (x) ρ (x )dx

whereρ ( x )is the probability distribution density function ofx . Substituting appropriate values:

⟨ x ⟩=∫−∞

∞

x ∙ 1σ √2 π

e−( x−μ )2

2σ 2 dx

Letz=x−μ:

⟨ x ⟩=∫−∞

∞

( z+μ ) ∙ 1σ √2 π

e− z2

2σ 2 dz= 1σ √2 π ∫−∞

∞

( z+μ ) ∙ e−z 2

2σ2 dz= 1σ √2π

∙∫−∞

∞

z ∙ e−z 2

2σ2 dz+ μσ √2π

∙∫−∞

∞

e−z2

2σ2 dz

¿0(Odd Function)+ μσ √2π

∙∫−∞

∞

e− z2

2σ 2dz

Substituting back:

⟨ x ⟩=μ∫−∞

∞ 1σ √2π

∙ e−( x−μ )2

2σ2 dx=¿μ∫−∞

∞

ρ ( x )dx=μ¿

c. Please calculate ⟨ x2 ⟩, ⟨ (x− ⟨ x ⟩ )2 ⟩ and ⟨ x2 ⟩− ⟨ x ⟩2

⟨ x2 ⟩=∫−∞

∞

x2∙ ρ ( x )dx=∫−∞

∞

x2∙ 1σ √2π

e−( x−μ )2

2σ2 dx

Let z=x−μ

σ anddz=dxσ :

⟨ x2 ⟩=∫−∞

∞

(μ+σz )2 ∙ 1√2π

e−z2

2 dz=∫−∞

∞

(μ2+2μσz+σ 2 z2 )∙ 1√2 π

e− z2

2 dz

¿ μ2∫−∞

∞1√2π

e− z2

2 dz+2μσ∫−∞

∞

z ∙ 1√2π

e−z2

2 dz+σ2∫−∞

∞

z2 ∙ 1√2π

e− z2

2 dz

Using:

ddb

I (b )=∫−∞

∞

y2 e−b y2dx= ddb √ π

b=−12b √ π

b

σ 2 1√2 π ∫−∞

∞

z2 ∙ e−z2

2 dz=σ2whereb=12

∴ ⟨x2 ⟩=μ2+σ2.

For⟨ (x− ⟨ x ⟩ )2 ⟩ and ⟨ x2 ⟩− ⟨ x ⟩2 :⟨ (x− ⟨ x ⟩ )2 ⟩=⟨ ( x2−2x ⟨ x ⟩+ ⟨ x ⟩2 ) ⟩= ⟨x2 ⟩−⟨ x ⟩2

Using above results:

⟨ x2 ⟩− ⟨ x ⟩2=μ2+σ2−μ2=σ2

5. We have a drunken sailor, who moves in 1D with the step size of L. At each time period, she could randomly take a step forward with probabilityP+¿=0.5¿, take a step backward with probability P−¿=0.25¿ or stay in position with the probabilityP0=0.25.

a. Find the sailor’s mean position after Nth step.b. Find the variance of the position after Nth step.c. What is her diffusion constant D?

a. Let Pk be the probability of taking steps of kL where k=0∨±1 , and L is the step length. Pk

is given by the following:

Pk={ 0.50.250.25

¿k=1¿k=0¿k=−1

Let the displacement of the jth step be ∆ x j=k j L and the initial position bex0=0. Let u be the

mean value ofk j:u=⟨k j ⟩=∑k

kPk=0.5×1+0.25×0+0.25×−1=0.25 and thus the mean

position of the walker after the Nth step would be:

⟨ xN ⟩=⟨x N−1 ⟩+ L ⟨k N ⟩= ⟨xN−1 ⟩+uL=NuL=0.25NLb. The variance is given by:

σ 2 (xN )=⟨ (x N− ⟨xN ⟩)2⟩= ⟨ (xN−1+kN L−NuL )2 ⟩=⟨( (xN−1−u (N−1 ) L )+(k N L−uL ))2⟩=⟨ (x N−1−u (N−1 )L )2+2 ⟨ (x N−1−u (N−1 )L ) (k N L−uL ) ⟩+L2 (k N−u )2 ⟩=⟨( xN−1−⟨x N−1 ⟩ )

2 ⟩+L2 ⟨ (kN− ⟨k N ⟩ )2 ⟩=Var (xN−1)+L2×Var (k )=N L2+Var (k )

ForVar (k ):

σ 2 (k )=⟨ (k− ⟨k ⟩ )2 ⟩=⟨k2 ⟩− ⟨k ⟩2=∑k

k 2Pk−∑k

kPk=(0.5×12+0.25×02+0.25×−12 )−0.25=0.5

c. The diffusion constant is given by

D= L2

2∆ t×Var (k )= L2

2∆ t×0.5

6. The chapter asserted that tiny objects stop moving at once when we stop pushing them.

a. Consider a bacterium, idealized as a sphere of radius 1m, propelling itself at 1m/s. At time zero, the bacterium suddenly stops swimming and coasts to a stop, following Newton’s Law of motion with the Stokes drag force. How far does it travel before it stops? Comment.

b. Our discussion of Brownian motion assumed that each random step was independent of the previous one; thus, for example, we neglected the possibility of a residual drift speed left over from the previous step. In the light of (a), would you say that this assumption is justified for a bacterium?

Assume viscosity of water, = 0.001Pas, density of bacteria is same as water, = 1000kg/m3

and low Reynolds number where Stokes law assumption is reasonable. For a spherical bacterium of radius 1m moving at 1m/s:

Mass of bacterium, m=ρV=1000× 43

π (1×10−6)3=4×10−15 kg

Using Newton’s Second Law:

F=m dvdt

The only force present is the viscous drag force which follows Stokes law: F=−6 πμrv

Doing a force balance:

−6 πμrv=m dvdt

−∫t=0

t 6 πμrm

dt=∫v=0

v dvv

−6πμrm

t=ln vv0

v=v0 ∙ e−6πμr

m t

Distance travelled until stationary:

v=dxdt

v0 ∙ e−6πμr

m t=dx

dt

∫t=0

t=∞

v0 ∙ e−6πμr

m tdt=∫

0

x

dx

m6πμr

v0=x

x= 4×10−15

6π (0.001)(10−6)10−6=2×10−13m

The bacterium travels about 20 pm before stopping almost instantaneously as the travelling distance is almost 7 times in magnitude smaller than the bacterium.

During the deceleration, we ignored any Brownian motion in the calculations as assumed that the flow due to propulsion is dominant.

D=kB T6 πμr=1.38×10−23×3006 π (0.001)(10−6)

=2×10−13m /s

From the calculation of the diffusion coefficient and assuming that diffusion occurs, it is noted that in one second the distance moved by the bacterium would be same as the coasting distance. Thus we cannot ignore Brownian motion in our calculations.

7.

a) The initial position is given byf /k while the new equilibrium position is given byf 1/k . Thus the displacement is:

∆ x=f−f 1

kThe work done against the external force would then be:

W=f 1∆ x=f 1( f−f 1k )

b) To find the maximal work, we need to optimize W, the work done:

∂W∂ f 1=( f−f 1

k )− f 1k=0

The solution to the above equation is f=2 f 1 and the corresponding work done is:

W=f 1( f− f 1k )= f

2 ( f−f2

k )= f 2

4 kThe useful work output is only half of what was stored in the spring.

c) To make the process more efficient, the difference between the external force f 1 and the original force f should be infinitesimally small for every value of x. This would ensure that all the stored energy in the spring is retrieved without losing any to friction.

8.

For the polymer bead tied to the membrane, its microstate can be described by its positionx and velocityv. Modelling it as a “spring”, the total energy would then be:

ET ( x , v )=12(m v2+k x2)

Using the Boltzmann distribution, the probability is given as follow where C is a constant:

P ( x , v )=C × e−12 (m v2+k x2)/ kBT

Using the fact that the sum of probabilities is 1:

PT ( x , v )=C∬−∞

∞

e−12 (mv2+k x2)/kB T

dvdx=C∫−∞

∞

e−12 mv2 /kB T

dv ∙∫−∞

∞

e−12 k x2 /kB T

dx=1

Using the following relation:

∫−∞

∞

e−b y2dy=√ πb

PT ( x , v )=C ∙√ 2 π kB Tm √ 2π kB T

k=C ∙

2π k BT√mk

=1

C= √mk2π k BT

∴P ( x , v )= √mk2π k BT

e−12 (m v2+k x2)/ kBT

For the average energy:

⟨ET ( x , v ) ⟩=∬−∞

∞

ET ( x , v ) ∙ P(x , v)dvdx= √mk2π k BT (∬−∞

∞ 12

m v2 ∙e−12 (m v2+k x2)/ kBT

dvdx+∬−∞

∞ 12

k x2 ∙ e−12 (mv2+k x2 )/ kB T

dvdx)= √mk2 π kB T (∫−∞

∞ 12

m v2∙ e−12

mv2/ kB Tdv ∙∫

−∞

∞

e−12 k x2/ kB T

dx+∫−∞

∞

e−12 mv2 / kBT

dv ∙∫−∞

∞ 12

k x2 ∙ e−12 k x2 /kB T

dx)= √mk2π kB T (√ 2π k B T

m∙ m2

∙ 12 √ π (2k BT )3

m3 +√ 2π k B Tm

∙ k2

∙ 12 √ π (2kB T )3

m3 )= k BT2+

kB T2=kB T

Therefore at room temperature, the mean energy of the bead is:

⟨ ET ⟩=kB T=1.38×10−23×298=4.14×10−21 JThe mean energy is related to the potential energy in a spring by the following equation:

⟨ ET ⟩=12

k ⟨x2 ⟩Equating the two equations and solving:

12

k ⟨ x2 ⟩=4.14×10−21

12

k (35×10−9 )2=4.14×10−21

∴ k=6.8×10−6N /m

9.

a) The hypothesis of the trap door suggests a two state system where the stiffness of the hair bundle can vary between two states. This is shown in the graph where the “active” hair bundle would behave like an ordinary spring with constant stiffness and thus we see a

constant gradient for displacements of −80<x←20 and20<x<80. For displacements of−20<x<20, similarly a constant stiffness value was observed but this time it is negative. Thus we can attribute the observations to the trap door hypothesis where the hair bundle can vary between two states.

b) The bump in the curve is rounded and not sharp likely due to cooperative effects of each hair in the bundle. Each hair would move individually until it becomes unstable, and then the entire bundle would jump to a thermodynamically stable position, resulting in a smooth curve and not a sharp jump.

c) It is likely that displacement of 0nm is an unstable equilibrium point and thus any random displacement from the surrounding fluid would cause it to move towards the other two stable equilibrium points of -20 and 20nm.

10. Evaluate:

∫−∞

∞

e− p2

2mkB T ∙ d3 p=∫−∞

∞

e−p x

2

2mkB T d px∫−∞

∞

e−p y

2

2m kBT d py∫−∞

∞

e− pz

2

2mkB T dz pz=∭−∞

∞

e−( px

2+ py2+pz

2 )2mkB T d pxd py d pz

11.

a) The osmotic pressure due to the solutes inside the bag is balanced by the osmotic pressure due to the 1mM salt solution outside. If 2mM salt solution is used, there will be a larger osmotic pressure inwards and this will result in the bags collapsing as water would seep out of the bag until the volume is reduced by half.

b) Assuming that the salt completely ionized and is monovalent.

AB(s )→ A+¿(aq)+B−¿( aq)¿ ¿

For a mole of salt, there would be two moles of ions. For glucose which does not ionize in water, two moles of glucose would be needed for every mole of salt to balance the osmotic pressure.

12.

a) The following relation can be used to calculate the average molar mass: p=cRT

28mmHg × 101325Pa760mmHg

=3733 Pa

c= pRT= 37338.314×303

=1.48mol/m3

M=60gL

× 1000 L1m3 × 1m3

1.48mol=40540 g

mol=40.5kDa

b) Flux across capillary walls - since blood plasma proteins have depleted by 10%, the osmotic pressure should reduce by 10%:

J=Lp ∆ p=7×10−6×373.3 Pa× 1atm101325Pa

=2.58×10−8cm/ s

Total flowrate:

Q=JA=2.58×10−8cm /s× 1m100cm

×250m2=6.45×10−8m3/ s

6.45×10−8m3

s× 1000L

m3 × 3600 shr

× 24hrday=5.57 L/day

Swollen bellies due to 5.57L of water going into the interstitial space per day and being retained in the body. 5.57L is almost equivalent to drinking 11 bottles of water.

13.

a) The depletion zone thickness is approximately equal to the size of small objects. If the radius of the protein is about 10nm, then the radius of the small object would probably be 0.1nm to 1nm.

b) The total free energy reduction is given by the following relation:

∆ F=ck BT ×2 R× A=Vc kB T

When surfaces stick, the overlapping regions have a volume of about:V=10nm ×10 μm2=0.1μm3

The free energy reduction is then given by:∆ F=Vc kB T=0.1μ m3×c ×k B T

For spherical proteins with volume fraction of 0.3:

0.3=c× 4 π3(10−8 )3

c ≈7×1022m−3

∴∆ F=0.1μm3×7×1022m−3×kB T=7000 k BT

This is significant!

14.

Using the Taylor Series forex:

ex ≈1+ x+ x2

2 !+ x3

3 !+…

Taking the first two terms of the series expansion:eV ≈1+V∧e−V ≈1−V

Substituting: d2Vd x2=−12

λD−2 [e−V−eV ]≈−1

2λD−2 [ (1−V )− (1+V ) ]≈−1

2λD−2 [−2V ]≈ λD

−2V

Solving the resulting linear second order differential equation:d2Vd x2−λD−2V=0

Solution is given as:

V=C1e√λD−2x+C2e

−√λD−2x=C1 e

x /λD+C2 e−x / λD

Using boundary conditionV ()=0:

V=C2 e−x / λD

Ifx≪ λD, potential will be very large but ifx≫ λD, there would be almost no potential. Thus charges on the surface in a salt solution would have an influence over a distance λD, but beyond this border the surface and the cloud are essentially neutral.

15.

3.5 H-bonds per molecule:

3.5×9k B T r=31.5×1.38×10−23×298=1.295×10−19 J /molecule

Energy for one mole of water:

1.295×10−19 J /molecule ×6.023×1023molecule /mol=78kJ /mol

Comparing with the actual value of 40.7kJ/mol our estimated value is fairly close.

16.

As temperature rises, there would be an increase in molecular vibrations and this causes water to ionise more, forming more hydrogen ions which reduces the pH. Another way to explain this phenomenon would be that when a water molecule ionises to form two ions, the entropy

increases. Since the chemical potential is defined as: μ=−T dSdN where in this case dS/dN is

positive, a higher temperature will cause the chemical potential to be more negative, favouring dissociation. At equilibrium:

K eq=¿¿

As pH is defined as−log10 ¿¿:

−log10 ¿¿

Thus an increase in T would lead to a decrease in pH, which results in an increase in the [H+]

17.

Concentration of acetic solution:1mol10 L=0.1M

Let x be the concentration of acetic acid that ionizes:

CH3COOH + H2O H3O+ + CH3COO-

[CH3COOH] [H3O+] [CH3COO-]Initial 0.1 0 0

Change −x +x +xEquilibrium 0.1− x x x

K a=10−4.76= x2

(0.1−x )

Solving:x=0.00131, thus the pH of the solution is − log10 [0.00131]=2.88

18.

a) Consider a spherical protein with radius 3nm and an electrical charge of 10e. Force balance:

Electrostatic forceon particle=Fluid drag on particle

qE=6 πμrv

v= qE6πμr= 10×1.6×10−19C ×200V /m6 π×0.001 kg/m∙ s×3×10−9m

=5.66×10−6m /s

b) ∆ v=(q2−q1 ) E6πμr

= 1.6×10−19C ×470V /m6 π×0.001 kg/m ∙s×3×10−9m

=1.33×10−6m /s=4.79×10−3m /h

If they ran the gel for 20hours, the difference in distance is about

4.79×10−3m /h×20h=0.1m=10cm. This would allow for clear separation of the two bands. However if we just consider the distance travelled by the normal hemoglobin in 20h, it would be

about 1m and thus the gel which they have made should be rather large. Running the gel for 20 hours consecutively would also generate a lot of heat but cooling system should have been in place for the experiment to work.

19.

For the vector function:

r (s )= ⟨ f (s ) , g(s)⟩=⟨Rcos( sR ) , Rsin( s

R )⟩ f ' (s )=−sin( sR ) , g ' (s )=cos( sR )The arc length is given by:

L=∫0

s

√[ f ' (s )]2+[ g '(s) ]2ds=∫0

s √[−sin( sR )]

2

+[cos( sR )]

2

ds=s

The unit tangent vector:

t̂ ( s )= r '(s)‖r '(s)‖

=⟨−sin( s

R ) ,cos ( sR )⟩1

=⟨−sin ( sR ) ,cos ( sR )⟩The derivative of the unit tangent vector:

t̂ ' ( s )=⟨−1R cos( sR ) ,− 1R sin( sR )⟩

The bend is defined as:

β (s )=d t̂ds=⟨−1R cos( s

R ) ,− 1R sin( sR )⟩Thus the elastic energy cost for a 90 bend is:

E=12

kB T ∫0

14 2πR

ds A β2= 12

k B TA ∫0

14 2πR

( d t̂ds )

2

ds=12

kB TA( 12 πR)( 1R2 )= πA4 R

k BT

20.

a) The observed optical rotation of is a linear function of the fraction of amino acids in an -helix. Thus in an all-or-none model where the chain exists as either a random coil or a helix, we would be able to observe the optical rotation for the alpha-helix. By varying the number of subunits N and temperature where only the alpha-helix is observed, we would be able to observe the change in optical rotation.

b) Letting the weight of the random coil be 1, the weight of the alpha helix conformation would then be:

e−(N ∆ G)/kB T=e− β(N ∆G )

Thus the probability of being in the alpha helix state:

e−β (N ∆G )

1+e−β (N ∆G)=e−N (∆ Ebond−T ∆ Sbond−T ∆S conf ) /kB T

1+e−N (∆ Ebond−T ∆Sbond−T ∆Sconf )/ kB T=e−N /kB (∆E bond

T−∆ Sbond−∆S conf )

1+e−N / kB(∆ Ebond

T −∆S bond−∆Sconf)

c) The sharpness of the transition would increase with N. Since this is an all-or-none model, only two states of the chain would be present at equilibrium where either all residues are in random coil or in the alpha helix state. In this state of maximal cooperativity, the higher the number of residues, the easier the transition would be. Thus if there were an infinite number of residues, we would observe a step function.

d) The hydrogen bond is being formed between the oxygen atom in the carbonyl group of monomer k and the hydrogen atom I the amide group on monomer k+4 when in the helical conformation. As such the last 2 residues are unable to form hydrogen bonding.

e) Tm should generally decrease with increasing N.

21.

a) The diffusion coefficient is:

D=kB T6 πμr= 1.38×10−23×3006π ×0.001×0.37×10−6

=5.936×10−13m2/s

For 2D diffusion:¿ x2≥4 Dt=4×5.936×10−13×30=7.12×10−11m2

¿ x≥√4Dt=√7.12×10−11=8.44×10−6m≈8.4 μmSince he calculated the root-mean-square displacement to be 7.84m, the measured value is within error of the predicted value.

b) The probability that the displacement of the particles has a magnitude between |r| and |r|+dr is:

P (|r|, t )dr= 14 πDt

exp (−r2

4Dt ) (2πrdr )= rdr2Dt

exp(−r2

4Dt )The number of particles within the first circle should therefore be:

N0 ∫0

7.84×10−6

4

P (|r|,t )dr=508 ∙ 12Dt ∫0

7.84× 10−6

4

r ∙exp (−r2

4Dt )dr ≈26.39

Counting the number of particles in the first circle, there are about 30 and thus the prediction is in line with the actual number. Similar calculations can be done for the larger rings by replacing the integration limits with d/4 and 2d/4 etc…