Question

One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the distance of each from the line are measured, and the sum of the squares of the distance divided by 100 is 36 in2.

a. Assuming only one-dimensional movement away from the line, calculate the diffusion coefficient of the jumping beans.
b. If the mean jump distance of a bean is equal to 0.1 inches, estimate the mean jump frequency of a bean.

Answer 1

a) Diffusion coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Step-by-step explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

`<r>^(2) = 2Dt`..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

`36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr`

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

`<r>^(2) = 2Dt`

`0.1^(2) = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds`

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz