Answer:
I was just working on this now
z² - 0.8z - 98
Step-by-step explanation:
To solve this problem, we first need to define some variables:
X: the average position of the atom at time t = Nr
p: the probability of transiting to the right
q: the probability of transiting to the left (q = 1-p)
N: the number of times the atom transits from one site to another
r: the time it takes for the atom to transit from one site to another
t: the total time for which we are calculating the average position
z: the mean square value of the atom's position at time t
With these definitions in mind, we can now proceed to solve the problem:
a. To calculate the average position of the atom at time t = Nr, we can use the following formula:
X = p*(X+1) + q*(X-1)
This formula states that the average position of the atom is equal to the probability of transiting to the right multiplied by the position one unit to the right, plus the probability of transiting to the left multiplied by the position one unit to the left.
Solving for X, we get:
X = (p-q)/(p+q)
Substituting p = 0.6 and q = 0.4, we get:
X = (0.6-0.4)/(0.6+0.4) = 0.2/1 = 0.2
This means that the average position of the atom at time t = Nr is 0.2 units to the right.
b. To calculate the mean square value (z-7)² at time t, we can use the following formula:
(z-7)² = p*((z+1)-7)² + q*((z-1)-7)²
This formula states that the mean square value is equal to the probability of transiting to the right multiplied by the square of the position one unit to the right minus 7, plus the probability of transiting to the left multiplied by the square of the position one unit to the left minus 7.
Solving for (z-7)², we get:
(z-7)² = p*(z+1-7)² + q*(z-1-7)²
= pz² + 2pz + p - 49 + qz² - 2qz - q - 49
= (p+q)*z² - 2(p-q)*z - 98
Substituting p = 0.6 and q = 0.4, we get:
(z-7)² = (0.6+0.4)*z² - 2(0.6-0.4)*z - 98
= z² - 0.8z - 98
This means that the mean square value (z-7)² at time t is a quadratic function of z.